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Balance Chemical Equations - Chemistry

  1. Which represents the correctly balanced equation (smallest whole-number coefficients)? Fe + O2 → Fe2O3
    Answer: 4Fe + 3O2 → 2Fe2O3 This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). The law of conservation of mass requires that atoms aren't created or destroyed in chemical reactions, only rearranged, so whatever atoms you start with (left side) must all appear in the products (right side)—same number, same types, just in different combinations. For Fe + O2 → Fe2O3, the unbalanced counts are left: 1 Fe, 2 O; right: 2 Fe, 3 O—so balance Fe by placing a 2 in front of Fe (left: 2 Fe, 2 O; right: 2 Fe, 3 O), then balance O by using fractions temporarily (3/2 O2 for 3 O on left), but multiply everything by 2 to get whole numbers: 4 Fe + 3 O2 → 2 Fe2O3 (left: 4 Fe, 6 O; right: 4 Fe, 6 O). Choice B correctly balances the equation with coefficients that produce equal atom counts—4 Fe and 6 O on both sides—using smallest whole numbers. Choice A fails because it balances Fe but leaves O imbalanced with 2 O on left and 3 O on right. The systematic balancing strategy: (1) Write the unbalanced equation with correct formulas; (2) Count atoms of each element on both sides; (3) Balance one element at a time, starting with metals like Fe, then oxygen; (4) Use fractions if needed and multiply to clear them; (5) Recount and verify all elements—keep practicing, it builds confidence!
  2. Balance the equation using the smallest whole-number coefficients: Ca(OH)2 + HCl → CaCl2 + H2O​​
    Answer: Ca(OH)2 + 2HCl → CaCl2 + 2H2O This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas. For Ca(OH)2 + HCl → CaCl2 + H2O, let's count: Left has 1 Ca, 2 O, 2 H (from OH), 1 H and 1 Cl (from HCl); Right has 1 Ca, 2 Cl, 2 H and 1 O. Calcium is balanced, but chlorine isn't (1 ≠ 2). We need 2HCl to provide 2 Cl atoms. With 2HCl, we now have 4 H total on the left (2 from Ca(OH)2 + 2 from 2HCl) and 2 O. On the right, CaCl2 has no H or O, so all must come from water—we need 2H2O to balance: Ca(OH)2 + 2HCl → CaCl2 + 2H2O. Verification: Ca: 1 = 1 ✓, O: 2 = 2 ✓, H: 4 = 4 ✓, Cl: 2 = 2 ✓. Choice B correctly shows this balanced equation. Choice A doesn't balance Cl or H, choice C creates a Ca imbalance, and choice D forgets to balance the water molecules. This acid-base neutralization shows calcium hydroxide (a base) reacting with hydrochloric acid to form salt and water.
  3. Balance the equation using the smallest whole-number coefficients: Ca(OH)2 + HCl → CaCl2 + H2O​
    Answer: Ca(OH)2 + 2HCl → CaCl2 + 2H2O This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing equations with polyatomic ions like OH⁻ (hydroxide) requires careful counting—Ca(OH)2 has one calcium and two OH groups, meaning 2 oxygens and 2 hydrogens total! For Ca(OH)2 + HCl → CaCl2 + H2O, count atoms: Left has Ca=1, O=2, H=3, Cl=1; Right has Ca=1, Cl=2, H=2, O=1. Calcium is balanced, but we need 2 chlorines on the right (in CaCl2), so we need 2HCl: Ca(OH)2 + 2HCl → CaCl2 + H2O. Recounting: Left has Ca=1, O=2, H=4, Cl=2; Right has Ca=1, Cl=2, H=2, O=1. Now chlorine balances but hydrogen and oxygen don't. Since Ca(OH)2 provides 2 OH groups that become water, we need 2H2O: Ca(OH)2 + 2HCl → CaCl2 + 2H2O. Final count: Ca(1=1), O(2=2), H(4=4), Cl(2=2)—all balanced! Choice B correctly shows this balanced neutralization reaction. Choice A doesn't balance hydrogen or oxygen, choice C unnecessarily doubles all coefficients, and choice D fails to account for all the water produced. This acid-base neutralization shows calcium hydroxide (a base) reacting with hydrochloric acid to produce salt and water—a pattern where H⁺ from acid combines with OH⁻ from base to form H2O!
  4. Balance the chemical equation using the smallest whole-number coefficients: Al + O2 → Al2O3
    Answer: 4Al + 3O2 → 2Al2O3 This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). The law of conservation of mass requires that atoms aren't created or destroyed in chemical reactions, only rearranged, so whatever atoms you start with (left side) must all appear in the products (right side)—same number, same types, just in different combinations. For the equation Al + O2 → Al2O3, start by counting atoms: left Al:1, O:2; right Al:2, O:3—imbalanced; balance Al by 2Al2O3 (right Al:4, O:6); now left Al:1 (need 4, so 4Al), O:2 (need 6, so 3O2); final check: Al:4=4, O:6=6. Choice B correctly balances the equation with coefficients that produce equal atom counts for all elements on both sides using smallest whole numbers: 4Al + 3O2 → 2Al2O3. For example, choice A fails with 2Al (Al:2 left= right:2) but O:2 left vs. 3 right—oxygen imbalance; use multiples to handle the odd number of oxygens. The systematic balancing strategy: (1) Write the unbalanced equation with correct formulas (check subscripts are right for each substance—this part doesn't change!). (2) Count atoms of each element on both sides—make a list: 'Left: 1 Al, 2 O. Right: 2 Al, 3 O' (shows imbalance). (3) Balance one element at a time: Start with the most complex molecule or an element appearing once on each side, then move to others. Place coefficients (whole numbers in front) to equalize counts. (4) Recount after each coefficient change (changing one coefficient affects multiple elements if molecule has multiple atom types). (5) Final check: count ALL elements—make sure every single element balances. Use smallest whole numbers (if all coefficients divisible by 2, divide them all). Balancing tips: (1) Save oxygen for last in combustion reactions (appears in multiple products—easier to balance after everything else). (2) Keep polyatomic ions together if they don't break apart (NO3⁻ in AgNO3 → NaNO3 stays as NO3⁻ unit, balance it as a unit). (3) If you get fractions, that's OK temporarily—just multiply all coefficients by the denominator at the end to clear fractions (1/2 O2 becomes 1 O2 if you multiply by 2). (4) Check your work by counting each element separately—don't assume it's balanced until you verify every element! Balancing takes practice but gets faster with pattern recognition.
  5. Balance the chemical equation using the smallest whole-number coefficients (do not change subscripts): CH4 + O2 → CO2 + H2O​
    Answer: CH4 + 2O2 → CO2 + 2H2O This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). The law of conservation of mass requires that atoms aren't created or destroyed in chemical reactions, only rearranged, so whatever atoms you start with (left side) must all appear in the products (right side)—same number, same types, just in different combinations. For the equation CH4 + O2 → CO2 + H2O, start by counting atoms: left has C:1, H:4, O:2; right has C:1, H:2, O:3—imbalanced in H and O; balance C (already 1=1), then H by placing 2 in front of H2O (now right H:4, O:4 from CO2's 2 + 2H2O's 2); now O left is 2 but right 4, so place 2 in front of O2 (left O:4); final check: C:1=1, H:4=4, O:4=4. Choice B correctly balances the equation with coefficients that produce equal atom counts for all elements on both sides using smallest whole numbers: CH4 + 2O2 → CO2 + 2H2O. For example, choice A fails because it has only 1 O2 (O:2 left) but right side O:4 (from CO2 + 2H2O), so oxygen is imbalanced—remember to adjust coefficients step by step to match all elements. The systematic balancing strategy: (1) Write the unbalanced equation with correct formulas (check subscripts are right for each substance—this part doesn't change!). (2) Count atoms of each element on both sides—make a list: 'Left: 1 C, 4 H, 2 O. Right: 1 C, 2 H, 3 O' (shows imbalance). (3) Balance one element at a time: Start with the most complex molecule or an element appearing once on each side, then move to others. Place coefficients (whole numbers in front) to equalize counts. (4) Recount after each coefficient change (changing one coefficient affects multiple elements if molecule has multiple atom types). (5) Final check: count ALL elements—make sure every single element balances. Use smallest whole numbers (if all coefficients divisible by 2, divide them all). Balancing tips: (1) Save oxygen for last in combustion reactions (appears in multiple products—easier to balance after everything else). (2) Keep polyatomic ions together if they don't break apart (NO3⁻ in AgNO3 → NaNO3 stays as NO3⁻ unit, balance it as a unit). (3) If you get fractions, that's OK temporarily—just multiply all coefficients by the denominator at the end to clear fractions (1/2 O2 becomes 1 O2 if you multiply by 2). (4) Check your work by counting each element separately—don't assume it's balanced until you verify every element! Balancing takes practice but gets faster with pattern recognition.
  6. Determine the correct coefficients to balance the equation using the smallest whole numbers: Zn + HCl → ZnCl2 + H2
    Answer: Zn + 2HCl → ZnCl2 + H2 This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). The law of conservation of mass requires that atoms aren't created or destroyed in chemical reactions, only rearranged, so whatever atoms you start with (left side) must all appear in the products (right side)—same number, same types, just in different combinations. For Zn + HCl → ZnCl2 + H2, unbalanced: left 1 Zn, 1 H, 1 Cl; right 1 Zn, 2 Cl, 2 H—so balance Cl and H by placing a 2 in front of HCl (left: 1 Zn, 2 H, 2 Cl), now matching right. Choice B correctly balances the equation with coefficients that produce equal atom counts—1 Zn, 2 H, 2 Cl on both sides—using smallest whole numbers. Choice A fails because it leaves H and Cl imbalanced with only 1 each on left but 2 on right. The systematic balancing strategy: (1) Write the unbalanced equation with correct formulas; (2) Count atoms of each element on both sides; (3) Balance elements in compounds first, like Cl here; (4) Recount after changes; (5) Ensure all balance with smallest numbers—you're doing great, these skills build quickly!
  7. What coefficients balance the equation using the smallest whole numbers? Ca(OH)2 + HCl → CaCl2 + H2O
    Answer: Ca(OH)2 + 2HCl → CaCl2 + 2H2O This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). For Ca(OH)2 + HCl → CaCl2 + H2O, count atoms carefully, treating OH as a unit: Left: Ca=1, O=2, H=2 (from OH) + 1 (from HCl)=3 total, Cl=1; Right: Ca=1, Cl=2, H=2, O=1. Calcium is balanced, but we need 2 Cl on the left, so use 2HCl. This gives Ca(OH)2 + 2HCl → CaCl2 + H2O. Recount: Left: Ca=1, O=2, H=4 (2 from OH + 2 from 2HCl), Cl=2; Right: Ca=1, Cl=2, H=2, O=1. We need 2H2O on the right to balance H and O. Final equation: Ca(OH)2 + 2HCl → CaCl2 + 2H2O. Choice A correctly shows this balance. Choice B doesn't balance chlorine (1≠2), choice C uses unnecessarily large coefficients, and choice D doesn't balance water molecules. This acid-base neutralization shows calcium hydroxide reacting with hydrochloric acid to form calcium chloride and water.
  8. Determine the correct coefficients to balance the equation (smallest whole numbers): Zn + HCl → ZnCl2 + H2
    Answer: Zn + 2HCl → ZnCl2 + H2 This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing requires finding coefficients that equalize atoms while keeping formulas intact—ZnCl2 means zinc chloride with one zinc bonded to two chlorines, and this formula cannot be changed! For Zn + HCl → ZnCl2 + H2, count atoms: Left has Zn=1, H=1, Cl=1; Right has Zn=1, H=2, Cl=2. Zinc is balanced (1=1), but both hydrogen and chlorine need work. Since ZnCl2 requires 2 chlorine atoms and HCl provides only 1, we need 2HCl: Zn + 2HCl → ZnCl2 + H2. Recounting gives Left: Zn=1, H=2, Cl=2; Right: Zn=1, H=2, Cl=2—everything balances perfectly! Choice B correctly shows this balanced equation with the smallest whole-number coefficients. Choice A fails to balance chlorine and hydrogen, choice C uses unnecessarily large coefficients (everything doubled), and choice D incorrectly changes the product formula to ZnCl, which would be zinc(I) chloride—a different compound! This single replacement reaction shows zinc metal displacing hydrogen from hydrochloric acid, producing hydrogen gas that bubbles out. The systematic approach recognizes that polyatomic groups or multiple atoms in a formula (like Cl2 in ZnCl2) often dictate the coefficients needed for their source molecules.
  9. Balance the chemical equation using the smallest whole-number coefficients: Zn + HCl → ZnCl2 + H2
    Answer: Zn + 2HCl → ZnCl2 + H2 This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). The law of conservation of mass requires that atoms aren't created or destroyed in chemical reactions, only rearranged, so whatever atoms you start with (left side) must all appear in the products (right side)—same number, same types, just in different combinations. For the equation Zn + HCl → ZnCl2 + H2, start by counting atoms: left Zn:1, H:1, Cl:1; right Zn:1, Cl:2, H:2—imbalanced in H and Cl; balance Cl by 2HCl (left H:2, Cl:2); now H left:2= right:2, Zn:1=1, Cl:2=2; final check all match. Choice B correctly balances the equation with coefficients that produce equal atom counts for all elements on both sides using smallest whole numbers: Zn + 2HCl → ZnCl2 + H2. For example, choice A fails with only 1 HCl (Cl:1 left vs. 2 right in ZnCl2), so chlorine is imbalanced—adding a coefficient to HCl fixes both H and Cl since they come together. The systematic balancing strategy: (1) Write the unbalanced equation with correct formulas (check subscripts are right for each substance—this part doesn't change!). (2) Count atoms of each element on both sides—make a list: 'Left: 1 Zn, 1 H, 1 Cl. Right: 1 Zn, 2 Cl, 2 H' (shows imbalance). (3) Balance one element at a time: Start with the most complex molecule or an element appearing once on each side, then move to others. Place coefficients (whole numbers in front) to equalize counts. (4) Recount after each coefficient change (changing one coefficient affects multiple elements if molecule has multiple atom types). (5) Final check: count ALL elements—make sure every single element balances. Use smallest whole numbers (if all coefficients divisible by 2, divide them all). Balancing tips: (1) Save oxygen for last in combustion reactions (appears in multiple products—easier to balance after everything else). (2) Keep polyatomic ions together if they don't break apart (NO3⁻ in AgNO3 → NaNO3 stays as NO3⁻ unit, balance it as a unit). (3) If you get fractions, that's OK temporarily—just multiply all coefficients by the denominator at the end to clear fractions (1/2 O2 becomes 1 O2 if you multiply by 2). (4) Check your work by counting each element separately—don't assume it's balanced until you verify every element! Balancing takes practice but gets faster with pattern recognition.
  10. Determine the correct coefficients to balance the equation (smallest whole numbers): C3H8 + O2 → CO2 + H2O
    Answer: C3H8 + 5O2 → 3CO2 + 4H2O This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). The law of conservation of mass requires that atoms aren't created or destroyed in chemical reactions, only rearranged, so whatever atoms you start with (left side) must all appear in the products (right side)—same number, same types, just in different combinations. For the equation C3H8 + O2 → CO2 + H2O, start by counting atoms: left C:3, H:8, O:2; right C:1, H:2, O:3—imbalanced; balance C by 3CO2 (right C:3, O:6); H by 4H2O (right H:8, O:4 more, total O:10); now O left 2 vs. 10, so 5O2 (O:10 left); final check: C:3=3, H:8=8, O:10=10. Choice D correctly balances the equation with coefficients that produce equal atom counts for all elements on both sides using smallest whole numbers: C3H8 + 5O2 → 3CO2 + 4H2O. For example, choice A fails with 4O2 (O:8 left) but right 3CO2 (O:6) +4H2O (O:4)=O:10, so oxygen doesn't match—try saving O for last and balance C and H first. The systematic balancing strategy: (1) Write the unbalanced equation with correct formulas (check subscripts are right for each substance—this part doesn't change!). (2) Count atoms of each element on both sides—make a list: 'Left: 3 C, 8 H, 2 O. Right: 1 C, 2 H, 3 O' (shows imbalance). (3) Balance one element at a time: Start with the most complex molecule or an element appearing once on each side, then move to others. Place coefficients (whole numbers in front) to equalize counts. (4) Recount after each coefficient change (changing one coefficient affects multiple elements if molecule has multiple atom types). (5) Final check: count ALL elements—make sure every single element balances. Use smallest whole numbers (if all coefficients divisible by 2, divide them all). Balancing tips: (1) Save oxygen for last in combustion reactions (appears in multiple products—easier to balance after everything else). (2) Keep polyatomic ions together if they don't break apart (NO3⁻ in AgNO3 → NaNO3 stays as NO3⁻ unit, balance it as a unit). (3) If you get fractions, that's OK temporarily—just multiply all coefficients by the denominator at the end to clear fractions (1/2 O2 becomes 1 O2 if you multiply by 2). (4) Check your work by counting each element separately—don't assume it's balanced until you verify every element! Balancing takes practice but gets faster with pattern recognition.
  11. Balance the equation using the smallest whole-number coefficients (do not change subscripts): Ca(OH)2 + HCl → CaCl2 + H2O
    Answer: Ca(OH)2 + 2HCl → CaCl2 + 2H2O This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). The law of conservation of mass requires that atoms aren't created or destroyed in chemical reactions, only rearranged, so whatever atoms you start with (left side) must all appear in the products (right side)—same number, same types, just in different combinations. For Ca(OH)2 + HCl → CaCl2 + H2O, unbalanced: left Ca1 O2 H2 (from OH2) + H1 Cl1 = Ca1 O2 H3 Cl1; right Ca1 Cl2 H2 O1—balance Cl with 2 HCl (left now Ca1 O2 H4 Cl2), then H and O with 2 H2O (right Ca1 Cl2 H4 O2). Choice A correctly balances the equation with coefficients that produce equal atom counts for all elements on both sides using smallest whole numbers: Ca1, O2, H4, Cl2. Choice B fails with 1 HCl, leaving H3 Cl1 left vs H2 Cl2 right—treat (OH)2 as a unit but balance atoms individually. The systematic balancing strategy: (1) Write equation. (2) Count including polyatomics. (3) Balance acid first. (4) Recount. (5) Check all—fantastic!
  12. Balance the decomposition reaction using the smallest whole-number coefficients: H2O2 → H2O + O2
    Answer: 2H2O2 → 2H2O + O2 This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). The law of conservation of mass requires that atoms aren't created or destroyed in chemical reactions, only rearranged, so whatever atoms you start with (left side) must all appear in the products (right side)—same number, same types, just in different combinations. For H2O2 → H2O + O2, unbalanced: left 2 H, 2 O; right 2 H, 1 O + 2 O = 3 O—balance by putting 2 in front of H2O2 (4 H, 4 O left), 2 in front of H2O (4 H, 2 O), and 1 O2 (2 O), total right 4 O. Choice B correctly balances the equation with coefficients that produce equal atom counts for all elements on both sides using smallest whole numbers: 4 H and 4 O. Choice A fails as is with 2 O left vs 3 O right—try doubling the reactant to match oxygen without fractions. The systematic balancing strategy: (1) Write equation. (2) Count. (3) Balance O first in decompositions. (4) Recount. (5) Verify— you're doing great, keep practicing!
  13. What is the coefficient of O2 when the equation is balanced with the smallest whole numbers? Al + O2 → Al2O3
    Answer: 3 This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). The law of conservation of mass requires that atoms aren't created or destroyed in chemical reactions, only rearranged, so whatever atoms you start with (left side) must all appear in the products (right side)—same number, same types, just in different combinations. For Al + O2 → Al2O3, unbalanced: left 1 Al, 2 O; right 2 Al, 3 O—put 2 in front of Al2O3 (4 Al, 6 O right), 4 in front of Al (4 Al left), 3 in front of O2 (6 O left), so coefficient of O2 is 3. Choice B correctly identifies the coefficient as 3, balancing to 4 Al and 6 O both sides with smallest whole numbers. Choice A with 2 would only give 4 O left for 2 O2, but right needs 6 O for 2 Al2O3—scale up to match odd oxygen. The systematic balancing strategy: (1) Write equation. (2) Count. (3) Balance Al first. (4) Recount. (5) Verify coefficient— you're mastering this!
  14. What coefficients balance the equation using the smallest whole numbers? H2 + O2 → H2O
    Answer: 2H2 + O2 → 2H2O This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). The law of conservation of mass requires that atoms aren't created or destroyed in chemical reactions, only rearranged, so whatever atoms you start with (left side) must all appear in the products (right side)—same number, same types, just in different combinations. For H2 + O2 → H2O, unbalanced counts: left 2 H, 2 O; right 2 H, 1 O—so balance O by placing a 2 in front of H2O (right: 4 H, 2 O; left: 2 H, 2 O), then balance H by placing a 2 in front of H2 (left: 4 H, 2 O), giving 2 H2 + O2 → 2 H2O. Choice A correctly balances the equation with coefficients that produce equal atom counts—4 H and 2 O on both sides—using smallest whole numbers. Choice B fails because it balances O incorrectly, resulting in 4 O on left but only 2 O on right with excess H on right. The systematic balancing strategy: (1) Write the unbalanced equation with correct formulas; (2) Count atoms of each element on both sides; (3) Balance one element at a time, like starting with H here; (4) Recount after each coefficient; (5) Use smallest whole numbers and double-check—great job tackling these, you're improving!
  15. What coefficients balance the equation using the smallest whole numbers? KClO3 → KCl + O2​
    Answer: 2KClO3 → 2KCl + 3O2 This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). The law of conservation of mass requires that atoms aren't created or destroyed in chemical reactions, only rearranged, so whatever atoms you start with (left side) must all appear in the products (right side)—same number, same types, just in different combinations. For the equation KClO3 → KCl + O2, start by counting atoms: left K:1, Cl:1, O:3; right K:1, Cl:1, O:2—imbalanced in O; to balance O (3 is odd, 2 even), use 2KClO3 (left K:2, Cl:2, O:6) and 2KCl (right K:2, Cl:2), then O right:2 from O2? Need 6, so 3O2 (O:6); final check: K:2=2, Cl:2=2, O:6=6. Choice B correctly balances the equation with coefficients that produce equal atom counts for all elements on both sides using smallest whole numbers: 2KClO3 → 2KCl + 3O2. For example, choice A fails with no coefficients (O:3 left vs. 2 right)—multiplying by 2 and adjusting O2 to 3/2 then clearing fractions by multiplying all by 2 works, but start with even multiples for odd oxygens. The systematic balancing strategy: (1) Write the unbalanced equation with correct formulas (check subscripts are right for each substance—this part doesn't change!). (2) Count atoms of each element on both sides—make a list: 'Left: 1 K, 1 Cl, 3 O. Right: 1 K, 1 Cl, 2 O' (shows imbalance). (3) Balance one element at a time: Start with the most complex molecule or an element appearing once on each side, then move to others. Place coefficients (whole numbers in front) to equalize counts. (4) Recount after each coefficient change (changing one coefficient affects multiple elements if molecule has multiple atom types). (5) Final check: count ALL elements—make sure every single element balances. Use smallest whole numbers (if all coefficients divisible by 2, divide them all). Balancing tips: (1) Save oxygen for last in combustion reactions (appears in multiple products—easier to balance after everything else). (2) Keep polyatomic ions together if they don't break apart (NO3⁻ in AgNO3 → NaNO3 stays as NO3⁻ unit, balance it as a unit). (3) If you get fractions, that's OK temporarily—just multiply all coefficients by the denominator at the end to clear fractions (1/2 O2 becomes 1 O2 if you multiply by 2). (4) Check your work by counting each element separately—don't assume it's balanced until you verify every element! Balancing takes practice but gets faster with pattern recognition.
  16. Balance the equation using the smallest whole-number coefficients: Na + Cl2 → NaCl
    Answer: 2Na + Cl2 → 2NaCl This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing requires recognizing that Cl2 is diatomic chlorine gas (two chlorine atoms bonded together) while NaCl is sodium chloride (table salt) with one sodium and one chlorine—these formulas represent how these substances actually exist! For Na + Cl2 → NaCl, count atoms: Left has Na=1, Cl=2; Right has Na=1, Cl=1. Sodium balances (1=1) but chlorine doesn't (2≠1). Since Cl2 provides 2 chlorine atoms, we need 2NaCl to use them both: Na + Cl2 → 2NaCl. Now Left: Na=1, Cl=2; Right: Na=2, Cl=2. Chlorine balances but sodium doesn't (1≠2), so we need 2Na: 2Na + Cl2 → 2NaCl. Final count: Na(2=2) and Cl(2=2)—perfectly balanced! Choice B correctly shows this balanced equation using the smallest whole-number coefficients. Choice A incorrectly suggests NaCl2 which doesn't exist (sodium only forms +1 ions), choice C adds unnecessary chlorine molecules, and choice D doubles the chlorine unnecessarily. The key insight is that diatomic molecules like Cl2 often require even numbers of products to balance. This vigorous reaction between sodium metal and chlorine gas produces the salt we use every day—showing how reactive elements combine to form stable compounds!
  17. Which represents the correctly balanced equation (smallest whole-number coefficients)? Fe + O2 → Fe2O3​
    Answer: 4Fe + 3O2 → 2Fe2O3 This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). The law of conservation of mass requires that atoms aren't created or destroyed in chemical reactions, only rearranged, so whatever atoms you start with (left side) must all appear in the products (right side)—same number, same types, just in different combinations. For the equation Fe + O2 → Fe2O3, start by counting atoms: left has Fe:1, O:2; right has Fe:2, O:3—imbalanced; balance Fe by placing 2 in front of Fe2O3 (right Fe:4, O:6); now left Fe:1 (need 4, so 4Fe), O:2 (need 6, so 3O2 since 3x2=6); final check: Fe:4=4, O:6=6. Choice B correctly balances the equation with coefficients that produce equal atom counts for all elements on both sides using smallest whole numbers: 4Fe + 3O2 → 2Fe2O3. For example, choice A fails because it has 2Fe (Fe:2 left) but right Fe:2 from Fe2O3, yet O:2 left vs. O:3 right—oxygen is imbalanced; always verify all elements after adjustments. The systematic balancing strategy: (1) Write the unbalanced equation with correct formulas (check subscripts are right for each substance—this part doesn't change!). (2) Count atoms of each element on both sides—make a list: 'Left: 1 Fe, 2 O. Right: 2 Fe, 3 O' (shows imbalance). (3) Balance one element at a time: Start with the most complex molecule or an element appearing once on each side, then move to others. Place coefficients (whole numbers in front) to equalize counts. (4) Recount after each coefficient change (changing one coefficient affects multiple elements if molecule has multiple atom types). (5) Final check: count ALL elements—make sure every single element balances. Use smallest whole numbers (if all coefficients divisible by 2, divide them all). Balancing tips: (1) Save oxygen for last in combustion reactions (appears in multiple products—easier to balance after everything else). (2) Keep polyatomic ions together if they don't break apart (NO3⁻ in AgNO3 → NaNO3 stays as NO3⁻ unit, balance it as a unit). (3) If you get fractions, that's OK temporarily—just multiply all coefficients by the denominator at the end to clear fractions (1/2 O2 becomes 1 O2 if you multiply by 2). (4) Check your work by counting each element separately—don't assume it's balanced until you verify every element! Balancing takes practice but gets faster with pattern recognition.
  18. Balance the chemical equation (smallest whole-number coefficients). Do not change subscripts: Na + Cl2 → NaCl
    Answer: 2Na + Cl2 → 2NaCl This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). For Na + Cl2 → NaCl, count atoms: Left: Na=1, Cl=2; Right: Na=1, Cl=1. Sodium is balanced but chlorine isn't (2≠1). Since Cl2 is diatomic (always comes as a pair), we need 2NaCl to use both chlorine atoms. This requires 2Na on the left as well. The balanced equation is 2Na + Cl2 → 2NaCl. Choice B correctly shows this balance with Na: 2=2 and Cl: 2=2. Choice A incorrectly writes NaCl2 (sodium doesn't form this compound - it's always NaCl with a 1:1 ratio), choice C adds unnecessary chlorine molecules, and choice D doesn't balance sodium (1≠2). This synthesis reaction shows sodium metal reacting with chlorine gas to form table salt (sodium chloride), demonstrating how reactive metals combine with halogens.
  19. What coefficient of H2O is needed to balance the equation using the smallest whole numbers? C2H6 + O2 → CO2 + H2O​​
    Answer: 6 This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas. For the combustion C2H6 + O2 → CO2 + H2O, we need to find the coefficient for H2O. Start by balancing carbon: C2H6 has 2 C, so we need 2CO2. For hydrogen: C2H6 has 6 H, and each H2O has 2 H, so we need 3H2O to get 6 H atoms. Now count oxygen on the right: 2CO2 gives 4 O, and 3H2O gives 3 O, totaling 7 O atoms. Since O2 provides pairs, we need 7/2 O2. To get whole numbers, multiply everything by 2: 2C2H6 + 7O2 → 4CO2 + 6H2O. The coefficient of H2O is 6 when using smallest whole numbers. Choice D correctly identifies that 6 is the coefficient of H2O in the balanced equation. Choices A (2), B (3), and C (4) would all leave hydrogen unbalanced—with 12 H atoms on the left (2×6), we need 6H2O to provide 12 H atoms on the right. Remember: in combustion reactions, balance C first, then H, then O last!
  20. Balance the equation (smallest whole-number coefficients): H2O2 → H2O + O2
    Answer: 2H2O2 → 2H2O + O2 This question tests your ability to balance chemical equations by adjusting coefficients so that the number of atoms of each element is equal on both sides, reflecting the law of conservation of mass. Balancing chemical equations means finding the right coefficients (the numbers you write in front of chemical formulas) that make the atom count equal on both sides of the arrow, while NEVER changing the subscripts inside formulas (those define what the substance is—changing them creates a different substance!). The law of conservation of mass requires that atoms aren't created or destroyed in chemical reactions, only rearranged, so whatever atoms you start with (left side) must all appear in the products (right side)—same number, same types, just in different combinations. For H2O2 → H2O + O2, unbalanced: left 2 H, 2 O; right 2 H, 3 O—so place a 2 in front of H2O2 (left: 4 H, 4 O) and a 2 in front of H2O (right: 4 H, 2 O + 2 O from O2 = 4 O), resulting in 2 H2O2 → 2 H2O + O2. Choice B correctly balances the equation with coefficients that produce equal atom counts—4 H and 4 O on both sides—using smallest whole numbers. Choice A fails because it doesn't adjust for the extra O on the right, leaving 2 O left and 3 O right. The systematic balancing strategy: (1) Write the unbalanced equation with correct formulas; (2) Count atoms of each element on both sides; (3) Balance one element at a time, starting with the decomposed compound; (4) Recount and adjust; (5) Verify with smallest numbers—keep going, balancing gets smoother with each try!

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