Introduction for Reactions - College Chemistry

When 25.0 mL of 0.400 M $HNO_3$ is neutralized by 0.250 M $NaOH$ , what volume of base is required?
Answer: 40.0 mL Acid-base neutralization problems require you to find the point where moles of acid equal moles of base. The key insight is that neutralization occurs when the number of hydrogen ions from the acid matches the number of hydroxide ions from the base. Start by calculating moles of acid: $n_{HNO_3} = (0.0250 \text{ L})(0.400 \text{ M}) = 0.0100 \text{ mol}$ Since both $HNO_3$ and $NaOH$ are monoprotic (each molecule provides one $H^+$ or $OH^-$ ), you need exactly 0.0100 mol of $NaOH$ for complete neutralization. Now find the volume of base needed: $V_{NaOH} = \frac{n_{NaOH}}{M_{NaOH}} = \frac{0.0100 \text{ mol}}{0.250 \text{ M}} = 0.0400 \text{ L} = 40.0 \text{ mL}$ This confirms answer C (40.0 mL) is correct. Looking at the wrong answers: A (15.6 mL) might result from incorrectly multiplying the acid volume by the ratio of concentrations (25.0 × 0.250/0.400), but this ignores stoichiometry. B (25.0 mL) assumes equal volumes are needed regardless of concentration differences—a common misconception. D (62.5 mL) could come from incorrectly dividing 25.0 by 0.400 instead of using proper mole calculations. Study tip: Always convert to moles first in neutralization problems. Remember the formula $M_1V_1 = M_2V_2$ works here because both acids and bases are monoprotic. For polyprotic species, you'd need to account for the number of $H^+$ or $OH^-$ ions per molecule.
In the oxidation-reduction reaction $MnO_4^- + Fe^{2+} \rightarrow Mn^{2+} + Fe^{3+}$ in acidic solution, which species undergoes reduction and what is the change in oxidation state?
Answer: $MnO_4^-$ undergoes reduction with a change of -5 in oxidation state When you encounter redox reactions, you need to track oxidation states to determine which species is oxidized (loses electrons) and which is reduced (gains electrons). The key is identifying how the oxidation states change from reactants to products. Let's analyze each species in this reaction. For $MnO_4^-$ , manganese has an oxidation state of +7 (since oxygen is -2, and the overall charge is -1: x + 4(-2) = -1, so x = +7). When it becomes $Mn^{2+}$ , the oxidation state drops from +7 to +2, a change of -5. This decrease in oxidation state means $MnO_4^-$ gains electrons and undergoes reduction. For iron, $Fe^{2+}$ has an oxidation state of +2, which increases to +3 in $Fe^{3+}$ . This +1 change means iron loses an electron and is oxidized. Choice B correctly identifies that $MnO_4^-$ undergoes reduction with a -5 change in oxidation state. Choice A incorrectly states that $Fe^{2+}$ undergoes reduction when it actually undergoes oxidation, and the change is +1, not -1. Choice C correctly identifies the +1 change for iron but wrongly calls it reduction instead of oxidation. Choice D correctly identifies $MnO_4^-$ as undergoing reduction but gives the wrong magnitude of change (-1 instead of -5). Remember: reduction means gaining electrons and decreasing oxidation state, while oxidation means losing electrons and increasing oxidation state. Always calculate the actual numerical change in oxidation states to verify your answer.
In the reaction $CaC_2(s) + 2H_2O(l) \rightarrow Ca(OH)_2(aq) + C_2H_2(g)$ , what volume of acetylene gas ( $C_2H_2$ ) at STP is produced from 16.0 g of calcium carbide ( $CaC_2$ )?
Answer: 5.59 L This problem tests your ability to perform stoichiometric calculations involving gas volumes at standard temperature and pressure (STP). When you see a balanced chemical equation with specific masses and need to find gas volumes, think moles and molar ratios. Start by finding the molar mass of $CaC_2$ : $40.08 + 2(12.01) = 64.10 \text{ g/mol}$ . Convert the given mass to moles: $\frac{16.0 \text{ g}}{64.10 \text{ g/mol}} = 0.2496 \text{ mol } CaC_2$ . From the balanced equation, the molar ratio shows that 1 mole of $CaC_2$ produces 1 mole of $C_2H_2$ . Therefore, 0.2496 mol of $CaC_2$ produces 0.2496 mol of $C_2H_2$ . At STP, one mole of any gas occupies 22.4 L. So: $0.2496 \text{ mol} \times 22.4 \text{ L/mol} = 5.59 \text{ L}$ . This confirms answer B is correct. Answer A (2.80 L) represents exactly half the correct volume, suggesting an error in the molar ratio or calculation. Answer C (11.2 L) is exactly half of 22.4 L, indicating someone might have incorrectly assumed 0.5 moles of product. Answer D (22.4 L) would be correct if you had exactly 1 mole of $CaC_2$ , but you're working with approximately 0.25 moles. Remember: always convert mass to moles first, apply the molar ratio from the balanced equation, then use the molar volume at STP (22.4 L/mol) for gas volume calculations.
A student mixes 100.0 mL of 0.200 M $HCl$ with 50.0 mL of 0.400 M $NaOH$ . What is the concentration of the excess reagent in the final solution?
Answer: 0.00 M (no excess) When you encounter acid-base neutralization problems, you need to determine whether the reactants are in stoichiometric proportions or if one is in excess. The balanced equation is $HCl + NaOH \rightarrow NaCl + H_2O$ , showing a 1:1 molar ratio. First, calculate the moles of each reactant. For $HCl$ : 0.1000 L × 0.200 M = 0.0200 mol. For $NaOH$ : 0.0500 L × 0.400 M = 0.0200 mol. Since both reactants have exactly 0.0200 moles and react in a 1:1 ratio, they completely neutralize each other with no excess reagent remaining. The final solution contains only the salt $NaCl$ and water, making the concentration of excess reagent 0.00 M, which is answer A. Answer B (0.033 M $HCl$ ) would be incorrect because it assumes $HCl$ is in excess, but the calculations show equal molar amounts. Answer C (0.067 M $NaOH$ ) similarly assumes $NaOH$ is in excess, which isn't the case. Answer D (0.100 M $HCl$ ) represents a significant miscalculation, perhaps confusing initial concentration with final concentration or forgetting about the neutralization reaction entirely. Study tip: Always start neutralization problems by calculating moles of each reactant, then compare to the stoichiometric ratio from the balanced equation. Only when moles are unequal will you have an excess reagent to calculate. Remember that "excess" means what's left over after complete reaction, not what you started with.
When balancing the equation $Al + O_2 \rightarrow Al_2O_3$ , what is the sum of all coefficients in the balanced equation?
Answer: 9 When you encounter chemical equation balancing problems, you're applying the law of conservation of mass—atoms can't be created or destroyed, so you need equal numbers of each type of atom on both sides of the equation. Start by counting atoms in the unbalanced equation: $Al + O_2 \rightarrow Al_2O_3$ . The left side has 1 aluminum and 2 oxygen atoms, while the right side has 2 aluminum and 3 oxygen atoms. To balance aluminum, you need 2 Al atoms on the left, giving you: $2Al + O_2 \rightarrow Al_2O_3$ . Now for oxygen: the right side has 3 oxygen atoms, but $O_2$ molecules come in pairs. To get 3 oxygen atoms on the left using $O_2$ molecules, you need to find the least common multiple of 2 and 3, which is 6. This means you need $3O_2$ (giving 6 oxygen atoms) on the left and $2Al_2O_3$ (also giving 6 oxygen atoms) on the right. The balanced equation becomes: $4Al + 3O_2 \rightarrow 2Al_2O_3$ . Adding all coefficients: 4 + 3 + 2 = 9. Choice A (6) likely comes from forgetting to double the $Al_2O_3$ coefficient. Choice B (7) might result from incorrectly balancing one element. Choice D (12) could come from using unnecessarily large coefficients that maintain balance but aren't in lowest terms. Always verify your balanced equation by counting atoms of each element on both sides, and remember to use the smallest whole-number coefficients possible.
In the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ , if 5.0 mol of $N_2$ and 12.0 mol of $H_2$ are allowed to react, how many moles of $NH_3$ can be produced?
Answer: 8.0 mol This is a limiting reagent problem, which requires you to determine which reactant will be completely consumed first and therefore limit the amount of product that can be formed. Start by examining the balanced equation: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ . The stoichiometric ratios tell you that 1 mole of $N_2$ reacts with 3 moles of $H_2$ to produce 2 moles of $NH_3$ . With 5.0 mol of $N_2$ available, you would need $5.0 \text{ mol } N_2 \times \frac{3 \text{ mol } H_2}{1 \text{ mol } N_2} = 15.0 \text{ mol } H_2$ . However, you only have 12.0 mol of $H_2$ , so hydrogen is the limiting reagent. Using the limiting reagent ( $H_2$ ) to calculate product: $12.0 \text{ mol } H_2 \times \frac{2 \text{ mol } NH_3}{3 \text{ mol } H_2} = 8.0 \text{ mol } NH_3$ Choice A (8.0 mol) is correct. Choice B (10.0 mol) likely comes from incorrectly assuming $N_2$ is limiting: $5.0 \times 2 = 10.0$ . Choice C (12.0 mol) represents a common error of thinking all the $H_2$ converts directly to $NH_3$ without considering stoichiometry. Choice D (15.0 mol) might result from incorrectly multiplying the excess $N_2$ by 3. Study tip: Always identify the limiting reagent first by calculating how much of one reactant you'd need based on the amount of the other. The reactant that "runs out first" determines your maximum product yield.
When aqueous solutions of $Ba(NO_3)_2$ and $Na_2SO_4$ are mixed, a white precipitate forms. What is the net ionic equation for this reaction?
Answer: $Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)$ When you encounter precipitation reactions, you need to identify what actually forms the precipitate and write the net ionic equation showing only the species that participate in forming the solid. First, let's understand what happens when $Ba(NO_3)_2$ and $Na_2SO_4$ mix. Both compounds dissociate completely in water, producing $Ba^{2+}$ , $NO_3^-$ , $Na^+$ , and $SO_4^{2-}$ ions. The white precipitate that forms is $BaSO_4$ , which is insoluble according to solubility rules (most sulfates are soluble except those of barium, lead, and mercury). The net ionic equation includes only the ions that directly participate in forming the precipitate. Since $Ba^{2+}$ and $SO_4^{2-}$ combine to form solid $BaSO_4$ , the net ionic equation is: $Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)$ . This is answer C. Answer A is incorrect because it doesn't balance properly and mixes ionic and molecular formulas inappropriately. Answer B shows the complete molecular equation, not the net ionic equation—it includes all compounds rather than just the participating ions. Answer D represents the complete ionic equation, showing all dissociated ions including the spectator ions ( $Na^+$ and $NO_3^-$ ) that don't participate in the precipitation. Remember: net ionic equations strip away spectator ions and show only the chemical change occurring. Always identify the precipitate first using solubility rules, then write the equation using only the ions that form that precipitate.
Which of the following best describes what happens when aqueous solutions of $FeCl_3$ and $NaOH$ are mixed?
Answer: A precipitation reaction occurs forming insoluble iron(III) hydroxide When you encounter a question about mixing ionic solutions, you need to predict what type of reaction will occur based on the chemical properties of the reactants and products. Let's analyze what happens when $FeCl_3$ and $NaOH$ mix. Iron(III) chloride dissociates into $Fe^{3+}$ and $Cl^-$ ions, while sodium hydroxide produces $Na^+$ and $OH^-$ ions. The key insight is recognizing that $Fe^{3+}$ and $OH^-$ can combine to form $Fe(OH)_3$ , which is insoluble in water. The reaction is: $FeCl_3 + 3NaOH \rightarrow Fe(OH)_3(s) + 3NaCl$ . This forms a precipitate, making answer C correct. Now let's examine why the other options are wrong. Option A suggests a redox reaction producing iron metal and chlorine gas, but neither $Fe^{3+}$ nor $Cl^-$ undergoes oxidation or reduction under these mild aqueous conditions. Option B describes acid-base neutralization, but while $NaOH$ is indeed a base, $FeCl_3$ isn't functioning as an acid here—it's simply providing $Fe^{3+}$ ions for precipitation. Option D proposes single displacement with sodium replacing iron, but this would require sodium metal, not $NaOH$ , and wouldn't occur anyway since iron is more reactive than sodium. Remember: when predicting reactions between ionic compounds, always consider solubility rules first. Metal hydroxides (except Group 1 and some Group 2) are typically insoluble, making precipitation reactions common when mixing metal cations with hydroxide ions.
A solution is prepared by dissolving 15.0 g of $NaCl$ in enough water to make 250.0 mL of solution. What is the molarity of this solution?
Answer: 1.03 M This question tests your ability to calculate molarity, one of the most fundamental concentration units in chemistry. When you see a problem asking for molarity with given mass and volume, immediately think: molarity equals moles of solute divided by liters of solution. To find the molarity, you need to convert the 15.0 g of NaCl to moles, then divide by the solution volume in liters. First, calculate moles of NaCl using its molar mass (58.44 g/mol): $\frac{15.0 \text{ g}}{58.44 \text{ g/mol}} = 0.257 \text{ mol}$ . Next, convert the volume to liters: 250.0 mL = 0.2500 L. Finally, calculate molarity: $M = \frac{0.257 \text{ mol}}{0.2500 \text{ L}} = 1.03 \text{ M}$ , confirming answer B. Choice A (0.600 M) results from incorrectly using the mass of NaCl (15.0) divided by the volume in mL (250), without any unit conversions. Choice C (2.56 M) comes from dividing grams by liters without converting to moles first (15.0 g ÷ 0.250 L). Choice D (4.11 M) represents dividing the mass by the molar mass incorrectly, likely multiplying instead of dividing somewhere in the calculation. Remember the molarity formula: M = mol/L. Always convert mass to moles using molar mass, and volume to liters. Write out each step clearly to avoid unit conversion errors, which are the most common mistakes in molarity problems.
When copper metal reacts with concentrated nitric acid, the products include copper(II) nitrate, nitrogen dioxide, and water. What is the coefficient of $HNO_3$ in the balanced equation?
Answer: 4 When you encounter redox reactions involving metals and acids, you need to balance both mass and charge by carefully tracking electron transfer and ensuring all atoms are accounted for. Let's work through this systematically. Copper metal ( $Cu$ ) reacts with concentrated nitric acid ( $HNO_3$ ) to produce copper(II) nitrate ( $Cu(NO_3)_2$ ), nitrogen dioxide ( $NO_2$ ), and water ( $H_2O$ ). The unbalanced equation is: $Cu + HNO_3 \rightarrow Cu(NO_3)_2 + NO_2 + H_2O$ Start by balancing copper: 1 $Cu$ on each side. Next, notice that copper(II) nitrate contains 2 nitrate ions, so you need at least 2 $HNO_3$ molecules just for that. However, some nitric acid is also reduced to form $NO_2$ . The balanced equation is: $Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$ Let's verify: 1 Cu, 4 N, 12 O, and 4 H on both sides. The charge is also balanced. Choice A (2) only accounts for the nitrate ions in the salt, ignoring the $NO_2$ formation. Choice C (6) overcounts the nitrogen species needed. Choice D (8) severely overcounts, likely from doubling the actual requirement unnecessarily. The correct answer is B (4). Study tip: In metal-acid reactions, remember that some acid molecules provide ions for salt formation while others undergo reduction. Always count both roles when determining coefficients, and verify your balanced equation by checking both mass and charge balance.
A solution contains 0.150 M $NaCl$ and 0.200 M $AgNO_3$ . When these solutions are mixed in equal volumes, which statement best describes the reaction outcome?
Answer: $NaCl$ is the limiting reactant, and 0.075 M $AgCl$ precipitates This question tests your understanding of limiting reactants and stoichiometry in precipitation reactions. When you see two ionic compounds that can form a precipitate, you need to determine which reactant will be completely consumed first. The balanced equation is: $NaCl + AgNO_3 \rightarrow AgCl(s) + NaNO_3$ . Since the stoichiometry is 1:1, you need equal moles of each reactant. When equal volumes are mixed, each solution is diluted by half. The new concentrations become: $NaCl$ : 0.075 M and $AgNO_3$ : 0.100 M. Since you need equal moles and $NaCl$ has the lower concentration, it's the limiting reactant. All 0.075 M of $NaCl$ will react to form 0.075 M $AgCl$ precipitate, with 0.025 M $AgNO_3$ remaining unreacted. Option A correctly identifies $NaCl$ as limiting and gives the right precipitate concentration. Option B incorrectly claims $AgNO_3$ is limiting—it's actually in excess. Option C makes the error of using the original $NaCl$ concentration (0.150 M) instead of the diluted concentration after mixing. Option D wrongly suggests both reactants are completely consumed, ignoring that $AgNO_3$ is in excess. Remember: when solutions are mixed, always account for dilution first, then apply stoichiometry. The limiting reactant is the one that produces fewer moles of product, and in precipitation problems, this determines how much solid forms.
In the half-reaction $MnO_4^- \rightarrow Mn^{2+}$ in acidic solution, how many electrons are involved and on which side of the equation do they appear?
Answer: 5 electrons on the left side When you encounter redox half-reactions, you need to balance both mass and charge by determining how many electrons are transferred and where they belong in the equation. To find the electrons in $MnO_4^- \rightarrow Mn^{2+}$ , start by identifying the oxidation states. In $MnO_4^-$ , manganese has an oxidation state of +7 (since oxygen is -2, and 4(-2) + 7 = -1 for the overall charge). In $Mn^{2+}$ , manganese clearly has an oxidation state of +2. The change from +7 to +2 means manganese gains 5 electrons, so this is a reduction half-reaction. Since electrons are gained during reduction, they must appear as reactants on the left side of the equation: $MnO_4^- + 5e^- \rightarrow Mn^{2+}$ . This confirms that 5 electrons appear on the left side. Looking at the wrong answers: A is incorrect because while it correctly places electrons on the left (reduction side), it miscounts the electron change as 3 instead of 5. B makes the same counting error and incorrectly places electrons on the right side, which would indicate oxidation rather than reduction. D correctly identifies 5 electrons but wrongly places them on the right side, again confusing this reduction with an oxidation. Remember this pattern: in reduction half-reactions, electrons always appear on the reactant side (left), while in oxidation half-reactions, they appear on the product side (right). Calculate the oxidation state change carefully to determine the exact number of electrons transferred.
In the reaction $2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$ , what mass of $KCl$ is produced when 4.90 g of $KClO_3$ decomposes completely?
Answer: 2.98 g This is a stoichiometry problem that requires you to convert from grams of reactant to grams of product using molar ratios. When you see a balanced chemical equation with mass calculations, think mole-to-mole conversions as your bridge between substances. Start by finding the molar masses: $KClO_3$ = 39.1 + 35.5 + 3(16.0) = 122.6 g/mol, and $KCl$ = 39.1 + 35.5 = 74.6 g/mol. Convert the given mass to moles: 4.90 g $KClO_3$ ÷ 122.6 g/mol = 0.0400 mol $KClO_3$ . From the balanced equation, the molar ratio is 2:2 (or 1:1) between $KClO_3$ and $KCl$ , so 0.0400 mol $KClO_3$ produces 0.0400 mol $KCl$ . Finally, convert to grams: 0.0400 mol $KCl$ × 74.6 g/mol = 2.98 g $KCl$ . Answer A (2.98 g) is correct. Answer B (3.65 g) likely results from incorrectly using the 2:3 ratio between $KClO_3$ and $O_2$ instead of the 2:2 ratio with $KCl$ . Answer C (4.90 g) assumes no mass change during the reaction, ignoring that oxygen gas is released. Answer D (5.95 g) suggests using an incorrect molar mass or ratio calculation. Remember the stoichiometry roadmap: grams → moles → mole ratio → moles → grams. Always check that your answer makes chemical sense—the product mass should be less than the reactant mass when gas is evolved.
Which of the following represents the correct oxidation states for all atoms in the compound $K_2Cr_2O_7$ ?
Answer: K = +1, Cr = +6, O = -2 When determining oxidation states in compounds, you need to apply the rule that the sum of all oxidation states must equal the overall charge of the compound. Since $K_2Cr_2O_7$ is neutral, all oxidation states must sum to zero. Start with the elements whose oxidation states you know with certainty. Potassium (K) is an alkali metal that always has a +1 oxidation state in compounds. Oxygen typically has a -2 oxidation state (except in peroxides). With two potassium atoms contributing +2 total and seven oxygen atoms contributing -14 total, you have: $2(+1) + 2(Cr) + 7(-2) = 0$ Solving for chromium: $+2 + 2(Cr) - 14 = 0$ , so $2(Cr) = +12$ , meaning each chromium atom has a +6 oxidation state. Looking at the wrong answers: Choice A incorrectly assigns Cr = +3, which would make the total charge $2(+1) + 2(+3) + 7(-2) = -6$ , not zero. Choice C mistakenly gives K = +2, but alkali metals never have +2 oxidation states in compounds. Choice D assigns Cr = +7, which would create a total charge of $2(+1) + 2(+7) + 7(-2) = +2$ , again not zero. The correct answer is B: K = +1, Cr = +6, O = -2. Study tip: Always start oxidation state problems by identifying elements with predictable oxidation states (alkali metals = +1, alkaline earth metals = +2, oxygen = -2), then use the charge balance rule to find the unknown oxidation states.
What is the oxidation state of sulfur in the compound $Na_2S_2O_3$ (sodium thiosulfate)?
Answer: +2 When you encounter oxidation state problems, you need to use the fact that the sum of all oxidation states in a neutral compound equals zero, and apply known oxidation states for common elements. In $Na_2S_2O_3$ , start with what you know: sodium (Na) always has an oxidation state of +1, and oxygen typically has an oxidation state of -2. Since there are 2 sodium atoms contributing +2 total and 3 oxygen atoms contributing -6 total, you have: 2(+1) + 2(S) + 3(-2) = 0. Solving this equation: +2 + 2S - 6 = 0, which gives 2S = +4, so S = +2. However, thiosulfate has a special structure where the two sulfur atoms are not equivalent - one sulfur is bonded to oxygen atoms (oxidation state +6) and the other is bonded to the first sulfur (oxidation state -2). The average oxidation state is (+6 + (-2))/2 = +2. Looking at the wrong answers: A) +1 would result from incorrectly assuming sulfur behaves like a typical metal. C) +3 might come from dividing the total positive charge needed (+4) by an incorrect assumption about sulfur distribution. D) +4 is the trap of assigning this total positive charge to just one sulfur atom instead of recognizing it's distributed between two sulfur atoms. Remember that in polyatomic ions like thiosulfate, atoms of the same element can have different oxidation states due to different bonding environments. Always consider the average when asked for "the" oxidation state of an element appearing multiple times.
A 0.500 g sample of an unknown metal reacts with excess hydrochloric acid to produce 0.560 L of hydrogen gas at STP. If the metal has a +2 oxidation state in the compound formed, what is the molar mass of the metal?
Answer: 20.0 g/mol When you encounter gas evolution reactions with metals, you're dealing with stoichiometry that connects moles of gas produced to moles of metal consumed. The key insight is using the ideal gas law and balanced chemical equations. Since the metal has a +2 oxidation state, the reaction is: $\text{M} + 2\text{HCl} \rightarrow \text{MCl}_2 + \text{H}_2$ . This shows a 1:1 mole ratio between metal and hydrogen gas produced. First, find moles of H₂ gas. At STP, one mole of any gas occupies 22.4 L, so: $\text{moles of H}_2 = \frac{0.560 \text{ L}}{22.4 \text{ L/mol}} = 0.0250 \text{ mol}$ Since the stoichiometry is 1:1, you also have 0.0250 mol of metal. The molar mass is: $\text{Molar mass} = \frac{0.500 \text{ g}}{0.0250 \text{ mol}} = 20.0 \text{ g/mol}$ Looking at the wrong answers: B) 24.3 g/mol results from incorrectly using 18.4 L/mol instead of 22.4 L/mol at STP. C) 40.0 g/mol comes from assuming a 1:2 ratio (thinking 2 moles of metal produce 1 mole of H₂, which reverses the actual stoichiometry). D) 48.6 g/mol combines both errors—wrong gas volume and wrong stoichiometry. The correct answer is A) 20.0 g/mol. Study tip: Always write the balanced equation first to establish the mole ratio, and remember that at STP, 1 mole of gas = 22.4 L. These gas stoichiometry problems follow a predictable pattern: gas volume → moles of gas → moles of reactant → molar mass.
When lead(II) nitrate solution is mixed with potassium iodide solution, a bright yellow precipitate forms. Which compound precipitates and what type of reaction occurs?
Answer: $PbI_2$ precipitates; this is a double displacement reaction When you see two ionic solutions being mixed and a precipitate forms, you're dealing with a precipitation reaction, which is a specific type of double displacement (metathesis) reaction. In these reactions, the cations and anions "swap partners" to form new compounds. Let's write the balanced equation for this reaction: $Pb(NO_3)_2(aq) + 2KI(aq) \rightarrow PbI_2(s) + 2KNO_3(aq)$ The bright yellow precipitate is lead(II) iodide ( $PbI_2$ ), which forms because it's insoluble in water. You can predict this using solubility rules: most iodides are soluble, but lead iodide is one of the notable exceptions. Potassium nitrate ( $KNO_3$ ) remains dissolved since all nitrates and all potassium compounds are soluble. Looking at the wrong answers: Choice A correctly identifies $PbI_2$ as the precipitate but incorrectly calls this a synthesis reaction. Synthesis reactions combine simpler substances to form more complex ones, not rearrange existing compounds. Choice B is wrong because $KNO_3$ is soluble and remains in solution, and this isn't a decomposition reaction where compounds break apart. Choice D incorrectly suggests $Pb(NO_3)_2$ precipitates (it's actually a reactant that's soluble) and misidentifies the reaction type—single displacement involves an element replacing another element in a compound. Study tip: When predicting precipitation reactions, always write the products by swapping ions, then use solubility rules to determine which product will be insoluble. The insoluble product is your precipitate, and these are always double displacement reactions.
Which statement best explains why the reaction $2Al(s) + 3CuSO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 3Cu(s)$ is classified as a single displacement reaction?
Answer: A more active metal displaces a less active metal from its compound When you encounter chemical reactions, identifying the reaction type helps predict products and understand the underlying chemistry. This question tests your ability to recognize single displacement reactions and understand the activity series of metals. In this reaction, aluminum metal is replacing copper in copper sulfate. The key insight is that aluminum is more reactive (more active) than copper, so it can displace copper from its compound. You can see this by noting that solid aluminum becomes part of a compound ( $Al_2(SO_4)_3$ ), while copper goes from being in a compound to existing as a free metal. This is the hallmark of single displacement: one free element displaces another element from its compound. Looking at the wrong answers: (A) describes decomposition reactions, where one compound breaks into multiple simpler substances - but here we have a metal and a compound as reactants, not just one compound breaking down. (B) describes double displacement reactions, where two compounds swap ions completely (like $AB + CD \rightarrow AD + CB$ ) - but this reaction involves a free metal, not two compounds. (D) describes acid-base neutralization reactions that produce salt and water - this reaction involves neither acids nor bases, and no water is formed. The correct answer is (C) because aluminum, being higher on the activity series than copper, displaces the less active copper from its sulfate compound. Study tip: For single displacement reactions, remember the pattern: Active metal + Metal compound → Less active metal + New compound. Always check if the free element is more reactive than the one in the compound.
When balancing the redox equation $Cr_2O_7^{2-} + Fe^{2+} + H^+ \rightarrow Cr^{3+} + Fe^{3+} + H_2O$ in acidic solution, what is the coefficient of $H^+$ in the balanced equation?
Answer: 14 When you encounter redox equations in acidic solution, you need to balance both mass and charge by systematically balancing each half-reaction separately, then combining them. Start with the reduction half-reaction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ . The dichromate ion gains 6 electrons (chromium goes from +6 to +3 oxidation state), requires 14 $H^+$ ions to balance the oxygen atoms, and produces 7 water molecules. The oxidation half-reaction is simpler: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ . Each iron ion loses one electron. To balance electrons, multiply the iron half-reaction by 6: $6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-$ Combining both half-reactions: $Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$ The coefficient of $H^+$ is 14, making answer D correct. Choice A (6) likely comes from incorrectly using the number of electrons transferred. Choice B (8) might result from miscounting oxygen atoms or forgetting that dichromate contains two chromium atoms. Choice C (12) could arise from arithmetic errors when balancing the chromium atoms or water molecules. Remember the systematic approach: separate into half-reactions, balance atoms other than H and O first, then add $H_2O$ for oxygen and $H^+$ for hydrogen in acidic solutions. Always verify your final equation balances both atoms and charge.
Which of the following reactions represents a double displacement (metathesis) reaction?
Answer: $AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)$ When you encounter questions about reaction types, focus on identifying the key structural changes happening to the compounds involved. Double displacement (metathesis) reactions have a very specific pattern: two ionic compounds exchange their ions to form two new compounds, following the general form AB + CD → AD + CB. Option C demonstrates this perfectly: $AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)$ . Here, the silver ion (Ag⁺) from silver nitrate combines with the chloride ion (Cl⁻) from sodium chloride, while the sodium ion (Na⁺) pairs with the nitrate ion (NO₃⁻). The ions have literally "switched partners" to form two new compounds. Option A represents a single displacement reaction, where zinc metal displaces copper from copper sulfate. Only one element is being replaced, not a complete ion exchange between two compounds. Option B shows a decomposition reaction, where one compound (calcium carbonate) breaks down into two simpler substances (calcium oxide and carbon dioxide). No ion exchange occurs here. Option D illustrates a synthesis or combination reaction, where two elements combine to form a single compound (water). This is the opposite of decomposition. To master reaction classification, memorize the double displacement pattern: look for two ionic compounds as reactants that form two new ionic compounds as products, with the positive and negative ions switching places. The telltale sign is often the formation of a precipitate, gas, or water as one of the products.

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