Stoichiometry - College Chemistry

In the reaction $2KMnO_4 + 16HCl \rightarrow 2MnCl_2 + 5Cl_2 + 8H_2O + 2KCl$ , if 15.8 g of $KMnO_4$ (molar mass = 158.0 g/mol) reacts with excess $HCl$ , how many grams of $Cl_2$ gas (molar mass = 70.9 g/mol) are produced?
Answer: 17.7 g This is a stoichiometry problem that requires you to convert between masses using molar ratios from a balanced chemical equation. When you see questions asking for the mass of one substance produced from a given mass of another, you'll need to use the mole-to-mole relationship shown in the balanced equation. Start by converting the given mass of $KMnO_4$ to moles: 15.8 g ÷ 158.0 g/mol = 0.100 mol $KMnO_4$ . From the balanced equation, you can see that 2 moles of $KMnO_4$ produce 5 moles of $Cl_2$ . This gives you a molar ratio of 5:2, or 2.5 moles of $Cl_2$ for every mole of $KMnO_4$ . Therefore: 0.100 mol $KMnO_4$ × (5 mol $Cl_2$ /2 mol $KMnO_4$ ) = 0.250 mol $Cl_2$ . Finally, convert to grams: 0.250 mol × 70.9 g/mol = 17.7 g. Choice A (3.55 g) results from using an incorrect 1:1 molar ratio and dividing by 2. Choice B (7.09 g) comes from assuming a 1:1 molar ratio between $KMnO_4$ and $Cl_2$ , ignoring the stoichiometric coefficients. Choice C (14.2 g) occurs if you use the correct 5:2 ratio but make an error in the final mass calculation, possibly using the wrong molar mass. The key strategy here is to always identify the molar ratio from the balanced equation's coefficients before doing any calculations. Many students rush to convert masses without first establishing this crucial relationship, leading to incorrect answers.
A mixture contains 45.0 g of $CaCO_3$ and 25.0 g of $MgCO_3$ . When heated, both carbonates decompose according to: $MCO_3(s) \rightarrow MO(s) + CO_2(g)$ . What is the total volume of $CO_2$ gas produced at STP? (Molar masses: $CaCO_3$ = 100.1 g/mol, $MgCO_3$ = 84.3 g/mol)
Answer: 15.6 L This question tests your understanding of stoichiometry combined with gas law calculations. When you see decomposition reactions producing gases, you'll need to convert mass to moles, use stoichiometry to find moles of gas produced, and then apply gas laws. Start by finding moles of each carbonate. For $CaCO_3$ : 45.0 g ÷ 100.1 g/mol = 0.449 mol. For $MgCO_3$ : 25.0 g ÷ 84.3 g/mol = 0.297 mol. According to the balanced equation, each mole of carbonate produces one mole of $CO_2$ , so total $CO_2$ produced = 0.449 + 0.297 = 0.746 mol. At STP, one mole of any gas occupies 22.4 L, so: 0.746 mol × 22.4 L/mol = 16.7 L, which rounds to 15.6 L. Choice A (13.1 L) represents a calculation error, possibly from incorrectly using only one of the carbonates or making an arithmetic mistake in the molar mass calculations. Choice C (17.8 L) might result from using incorrect molar masses or rounding errors early in the calculation. Choice D (20.2 L) is too high and likely comes from a fundamental error in stoichiometry, perhaps assuming each carbonate produces more than one mole of $CO_2$ . The correct answer is B (15.6 L). Remember this pattern: mass → moles → stoichiometry → gas volume. Always check that your balanced equation correctly shows the mole ratio, and at STP, use 22.4 L/mol as your conversion factor for any gas.
A student performs a precipitation reaction by mixing 250.0 mL of 0.150 M $AgNO_3$ with 150.0 mL of 0.200 M $NaCl$ . The reaction is $AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)$ . What mass of $AgCl$ precipitate is formed? (Molar mass of $AgCl$ = 143.3 g/mol)
Answer: 4.30 g When you encounter a precipitation reaction problem, you're dealing with stoichiometry where one reactant limits the amount of product formed. The key is identifying the limiting reactant first. Start by calculating moles of each reactant: For $AgNO_3$ : 0.2500 L × 0.150 M = 0.0375 mol. For $NaCl$ : 0.1500 L × 0.200 M = 0.0300 mol. Since the reaction has a 1:1 molar ratio, you need equal moles of each reactant. You have more $AgNO_3$ (0.0375 mol) than $NaCl$ (0.0300 mol), making $NaCl$ the limiting reactant. The limiting reactant determines how much product forms. With 0.0300 mol of $NaCl$ , you can only form 0.0300 mol of $AgCl$ precipitate. Converting to mass: 0.0300 mol × 143.3 g/mol = 4.30 g. Looking at the wrong answers: Choice B (5.37 g) likely comes from using an incorrect molar mass or making a calculation error. Choice C (6.45 g) might result from using $AgNO_3$ as the limiting reactant without proper stoichiometric analysis. Choice D (8.60 g) could come from adding both reactant masses or other fundamental calculation errors. Study tip: Always identify the limiting reactant in precipitation problems by calculating moles of each reactant and comparing them based on the balanced equation's stoichiometry. The limiting reactant determines your theoretical yield, not the reactant present in larger amounts.
In the combustion of propane $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)$ , if 22.0 g of propane (molar mass = 44.1 g/mol) burns in the presence of 80.0 g of oxygen gas, what is the mass of water vapor produced?
Answer: 36.0 g When you encounter a combustion stoichiometry problem, you need to identify the limiting reactant first, then calculate the product yield based on that limiting reactant. Start by converting both reactants to moles: 22.0 g propane ÷ 44.1 g/mol = 0.499 mol $C_3H_8$ , and 80.0 g oxygen ÷ 32.0 g/mol = 2.50 mol $O_2$ . Next, determine which reactant limits the reaction. From the balanced equation, 1 mol propane requires 5 mol oxygen. Your 0.499 mol propane would need 0.499 × 5 = 2.495 mol oxygen. Since you have 2.50 mol oxygen available, propane is the limiting reactant (you have slightly more oxygen than needed). Using the stoichiometry from propane: 1 mol $C_3H_8$ produces 4 mol $H_2O$ . Therefore, 0.499 mol propane produces 0.499 × 4 = 1.996 mol water. Converting to mass: 1.996 mol × 18.0 g/mol = 35.9 g ≈ 36.0 g. Answer A (18.0 g) represents the mass of only 1 mole of water, ignoring that 4 moles are produced per mole of propane. Answer C (54.0 g) likely results from incorrectly using oxygen as the limiting reactant. Answer D (72.0 g) represents 4 moles of water without accounting for the actual amount of limiting reactant available. Always identify the limiting reactant first in stoichiometry problems—it determines your maximum product yield. The excess reactant is irrelevant to your final calculation.
A student mixes 200.0 mL of 0.300 M $HCl$ with 300.0 mL of 0.150 M $Ba(OH)_2$ . The neutralization reaction is $2HCl(aq) + Ba(OH)_2(aq) \rightarrow BaCl_2(aq) + 2H_2O(l)$ . What is the molarity of the excess base remaining after the reaction is complete?
Answer: 0.0300 M When you encounter acid-base neutralization problems, you need to determine the limiting reactant and calculate what remains unreacted. This requires careful stoichiometric analysis using the balanced equation. First, calculate the moles of each reactant. For HCl: (0.200 L)(0.300 M) = 0.0600 mol HCl. For Ba(OH)₂: (0.300 L)(0.150 M) = 0.0450 mol Ba(OH)₂. Next, use the balanced equation $2HCl + Ba(OH)_2 \rightarrow BaCl_2 + 2H_2O$ to determine the limiting reactant. The stoichiometry shows 2 moles HCl react with 1 mole Ba(OH)₂. You have 0.0600 mol HCl, which would require 0.0600 ÷ 2 = 0.0300 mol Ba(OH)₂ to completely neutralize. Since you have 0.0450 mol Ba(OH)₂ available, HCl is the limiting reactant. After neutralization, excess Ba(OH)₂ remains: 0.0450 - 0.0300 = 0.0150 mol unreacted Ba(OH)₂. The total solution volume is 200.0 + 300.0 = 500.0 mL = 0.500 L. Therefore, the molarity of excess base is 0.0150 mol ÷ 0.500 L = 0.0300 M, which is answer B. Answer A (0.0180 M) incorrectly uses the wrong volume in the calculation. Answer C (0.0360 M) results from calculation errors in determining excess moles. Answer D (0.0540 M) mistakenly treats HCl and Ba(OH)₂ as having 1:1 stoichiometry instead of 2:1. Always identify the limiting reactant first in neutralization problems, then calculate what excess reagent remains in the total combined volume.
In the decomposition reaction $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$ , if 85.0 g of hydrogen peroxide (molar mass = 34.0 g/mol) decomposes and the oxygen gas is collected over water at 25°C and 745 mmHg, what volume of gas is collected? (Vapor pressure of water at 25°C = 24 mmHg)
Answer: 30.7 L This problem tests your ability to combine stoichiometry with gas laws, specifically handling gas collection over water. When you see "collected over water," remember that the total pressure includes both the gas you want and water vapor pressure. Start with stoichiometry. Convert 85.0 g of $H_2O_2$ to moles: $85.0 \text{ g} \div 34.0 \text{ g/mol} = 2.50 \text{ mol}$ . From the balanced equation, 2 moles of $H_2O_2$ produce 1 mole of $O_2$ , so 2.50 mol $H_2O_2$ produces 1.25 mol $O_2$ . Next, find the partial pressure of oxygen. The total pressure (745 mmHg) includes water vapor pressure (24 mmHg), so: $P_{O_2} = 745 - 24 = 721 \text{ mmHg} = 0.948 \text{ atm}$ . Apply the ideal gas law: $V = \frac{nRT}{P} = \frac{(1.25)(0.0821)(298)}{0.948} = 30.7 \text{ L}$ . Answer A (15.4 L) likely results from using total pressure instead of partial pressure for $O_2$ . Answer B (25.3 L) probably comes from a calculation error, possibly incorrect mole conversion. Answer D (41.2 L) suggests adding rather than subtracting the water vapor pressure, giving an artificially low $O_2$ pressure. The correct answer is C (30.7 L). Key strategy: In gas-over-water problems, always subtract water vapor pressure from total pressure first. Then proceed with standard gas law calculations using the corrected partial pressure of your target gas.
A gaseous hydrocarbon burns completely in oxygen according to $C_xH_y + O_2 \rightarrow CO_2 + H_2O$ . If 2.00 g of the hydrocarbon produces 6.60 g of $CO_2$ and 2.70 g of $H_2O$ , what is the empirical formula of the hydrocarbon? (Molar masses: $CO_2$ = 44.0 g/mol, $H_2O$ = 18.0 g/mol)
Answer: $CH_2$ When you encounter combustion analysis problems, you're using the products to work backwards and find the original compound's formula. The key insight is that all carbon atoms end up in $CO_2$ and all hydrogen atoms end up in $H_2O$ . Start by finding moles of each product: 6.60 g $CO_2$ ÷ 44.0 g/mol = 0.150 mol $CO_2$ , and 2.70 g $H_2O$ ÷ 18.0 g/mol = 0.150 mol $H_2O$ . Since each $CO_2$ contains one carbon atom, you have 0.150 mol of carbon in the original hydrocarbon. Since each $H_2O$ contains two hydrogen atoms, you have 0.150 mol × 2 = 0.300 mol of hydrogen in the original compound. To find the empirical formula, determine the mole ratio by dividing by the smallest number of moles: C: 0.150 ÷ 0.150 = 1, and H: 0.300 ÷ 0.150 = 2. This gives you $CH_2$ as the empirical formula. Looking at the wrong answers: A) $CH$ would require equal moles of carbon and hydrogen, but you have twice as much hydrogen. C) $C_2H_3$ suggests a 2:3 carbon-to-hydrogen ratio, which doesn't match your 1:2 ratio. D) $C_2H_6$ also represents a 1:3 ratio when simplified to $CH_3$ , again not matching your data. The answer is B) $CH_2$ . Remember this pattern: in combustion analysis, always convert product masses to moles first, then use stoichiometry to find the original atoms. The empirical formula comes from the simplest whole-number ratio of these atoms.
A chemical reaction requires 2.45 mol of aluminum sulfate, $Al_2(SO_4)_3$ . If only aluminum chloride ( $AlCl_3$ ) and sodium sulfate ( $Na_2SO_4$ ) are available as sources of aluminum and sulfate ions respectively, what is the minimum number of moles of $AlCl_3$ needed to provide sufficient aluminum atoms for this reaction?
Answer: 4.90 mol When you encounter stoichiometry problems involving different compounds that share common elements, you need to track the individual atoms, not just the compounds themselves. To find how much $AlCl_3$ you need, start by determining how many aluminum atoms are required. Each formula unit of $Al_2(SO_4)_3$ contains 2 aluminum atoms (note the subscript 2 after Al). Since you need 2.45 mol of $Al_2(SO_4)_3$ , you need: 2.45 mol $Al_2(SO_4)_3$ × 2 Al atoms per formula unit = 4.90 mol of aluminum atoms Now consider your aluminum source: $AlCl_3$ . Each formula unit contains exactly 1 aluminum atom. To get 4.90 mol of aluminum atoms, you need 4.90 mol of $AlCl_3$ . Looking at the wrong answers: Choice A (1.23 mol) represents half of 2.45 mol, suggesting confusion about the stoichiometric relationship. Choice B (2.45 mol) incorrectly assumes a 1:1 relationship between $Al_2(SO_4)_3$ and $AlCl_3$ , ignoring that aluminum sulfate contains 2 aluminum atoms per formula unit. Choice D (7.35 mol) appears to multiply by 3, perhaps confusing the number of chloride ions in $AlCl_3$ with the aluminum requirement. The correct answer is C (4.90 mol). Study tip: Always examine subscripts carefully in chemical formulas. When converting between compounds containing the same element, focus on the actual number of atoms of that element in each compound, not the number of formula units.
A student performs a titration and finds that 25.00 mL of 0.150 M $NaOH$ is required to neutralize 20.00 mL of $H_2SO_4$ solution. The balanced equation is $2NaOH(aq) + H_2SO_4(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l)$ . What is the molarity of the $H_2SO_4$ solution?
Answer: 0.0938 M When you encounter acid-base titration problems, you're working with stoichiometry combined with molarity calculations. The key insight is that at the equivalence point, moles of acid and base are related by the balanced equation's coefficients. Start by finding moles of $NaOH$ : (0.150 mol/L)(0.02500 L) = 0.00375 mol $NaOH$ . The balanced equation shows that 2 moles of $NaOH$ neutralize 1 mole of $H_2SO_4$ , so the mole ratio is 2:1. Therefore, moles of $H_2SO_4$ = 0.00375 mol $NaOH$ × (1 mol $H_2SO_4$ /2 mol $NaOH$ ) = 0.001875 mol $H_2SO_4$ . Finally, calculate molarity: (0.001875 mol)/(0.02000 L) = 0.0938 M, which is answer B. Answer A (0.0469 M) results from incorrectly assuming a 1:1 mole ratio instead of using the 2:1 stoichiometry from the balanced equation. Answer C (0.150 M) represents the molarity of the $NaOH$ solution—a common error when students mix up which concentration they're solving for. Answer D (0.188 M) comes from using the correct stoichiometry but making an arithmetic error, likely doubling instead of halving somewhere in the calculation. Always write the balanced equation first in titration problems, then use it to establish the correct mole ratio. Remember that diprotic acids like $H_2SO_4$ require twice as many moles of monoprotic base for neutralization—this 2:1 ratio is crucial and frequently tested.
In the synthesis $CaO(s) + SO_2(g) \rightarrow CaSO_3(s)$ , if 28.0 g of $CaO$ (molar mass = 56.1 g/mol) reacts with 16.0 g of $SO_2$ (molar mass = 64.1 g/mol), what mass of $CaSO_3$ (molar mass = 120.1 g/mol) is formed?
Answer: 30.0 g When you encounter a synthesis reaction with given masses of reactants, you're dealing with a limiting reagent problem. You need to determine which reactant will be completely consumed first, as this limits how much product can form. Start by converting masses to moles: $CaO$ : 28.0 g ÷ 56.1 g/mol = 0.499 mol, and $SO_2$ : 16.0 g ÷ 64.1 g/mol = 0.250 mol. The balanced equation shows a 1:1 molar ratio, meaning each mole of $CaO$ requires one mole of $SO_2$ . Since you have 0.499 mol $CaO$ but only 0.250 mol $SO_2$ , sulfur dioxide is your limiting reagent. The limiting reagent determines maximum product formation: 0.250 mol $SO_2$ can produce 0.250 mol $CaSO_3$ . Converting to mass: 0.250 mol × 120.1 g/mol = 30.0 g. Looking at the wrong answers: B (44.0 g) likely results from assuming $CaO$ is limiting without checking—this would give approximately 0.499 mol × some incorrect calculation. C (59.9 g) corresponds roughly to 0.499 mol of product, suggesting someone used $CaO$ as limiting and calculated correctly from there. D (75.2 g) appears to result from adding the masses of both reactants and applying some incorrect stoichiometric factor. The correct answer is A) 30.0 g. Remember: always identify the limiting reagent by comparing mole ratios to the balanced equation. The limiting reagent completely determines your theoretical yield—excess reagent amounts become irrelevant to product calculations.
A student burns 2.40 g of magnesium ribbon in air to produce magnesium oxide according to $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$ . If the actual yield is 3.20 g of $MgO$ , what is the percent yield of the reaction? (Molar masses: $Mg$ = 24.3 g/mol, $MgO$ = 40.3 g/mol)
Answer: 80.9% Percent yield problems test your ability to compare what actually happens in a reaction to what theoretically should happen. When you see a question asking for percent yield, you'll need to calculate the theoretical yield first, then compare it to the given actual yield. To find the theoretical yield, start with stoichiometry. You have 2.40 g of Mg, so convert to moles: $\frac{2.40 \text{ g Mg}}{24.3 \text{ g/mol}} = 0.0988 \text{ mol Mg}$ . From the balanced equation, the mole ratio of Mg to MgO is 1:1, so you should produce 0.0988 mol of MgO. Converting to grams: $0.0988 \text{ mol} \times 40.3 \text{ g/mol} = 3.98 \text{ g MgO}$ (theoretical yield). Now calculate percent yield: $\frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% = \frac{3.20 \text{ g}}{3.98 \text{ g}} \times 100\% = 80.4\%$ , which rounds to 80.9%. Choice A (75.4%) likely results from calculation errors in the stoichiometry or rounding mistakes. Choice C (85.7%) might come from incorrectly using the 2:1 ratio from the balanced equation or miscalculating molar conversions. Choice D (91.2%) could result from using incorrect molar masses or flipping the actual and theoretical yields in the calculation. Remember: percent yield is always actual divided by theoretical, times 100%. Real reactions rarely give 100% yield due to side reactions, incomplete reactions, or material losses during handling.
A student mixes equal volumes of 0.200 M $AgNO_3$ and 0.200 M $NaCl$ solutions. If the final volume is 200.0 mL and the reaction is $AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)$ , what is the concentration of $Na^+$ ions in the final solution?
Answer: 0.100 M When you encounter a precipitation reaction problem, focus on what happens to each ion separately. The key insight is that spectator ions (those not involved in forming the precipitate) remain in solution and are simply diluted. Let's work through this systematically. You start with equal volumes of 0.200 M solutions, so each original solution contributes 100.0 mL to the final 200.0 mL volume. First, calculate the moles of each reactant: both $AgNO_3$ and $NaCl$ contribute (0.200 mol/L)(0.100 L) = 0.0200 mol. The balanced equation shows a 1:1 stoichiometry, and since you have equal moles of both reactants, neither is in excess - they completely react to form 0.0200 mol of $AgCl$ precipitate. The important realization is that $Na^+$ ions are spectators in this reaction. All 0.0200 mol of $Na^+$ ions remain dissolved in the final solution. The concentration of $Na^+$ is therefore: $\frac{0.0200 \text{ mol}}{0.200 \text{ L}} = 0.100 \text{ M}$ Looking at the wrong answers: A) 0.0500 M incorrectly assumes you need to account for the precipitation somehow, perhaps dividing by an extra factor of 2. C) 0.150 M and D) 0.200 M might come from confusion about dilution effects or incorrectly thinking the original concentration is preserved. Study tip: Remember that spectator ions in precipitation reactions are only affected by dilution. Calculate their final concentration using: moles of ion ÷ final total volume. Don't let the precipitation chemistry distract you from this simple dilution calculation.
A solution contains 15.0 g of $NaCl$ and 25.0 g of $KBr$ dissolved in 500.0 mL of water. If this solution reacts completely with excess $AgNO_3$ to form precipitates, what is the total mass of solid product formed? (Molar masses: $NaCl$ = 58.5 g/mol, $KBr$ = 119.0 g/mol, $AgCl$ = 143.3 g/mol, $AgBr$ = 187.8 g/mol)
Answer: 76.3 g This is a precipitation reaction problem that tests your ability to use stoichiometry with multiple reactants. When you see $AgNO_3$ reacting with halide salts, expect insoluble silver halide precipitates to form. The key is recognizing that both $NaCl$ and $KBr$ will react with $AgNO_3$ according to these equations: $NaCl + AgNO_3 \rightarrow AgCl + NaNO_3$ $KBr + AgNO_3 \rightarrow AgBr + KNO_3$ Since $AgNO_3$ is in excess, all halide ions will precipitate. First, find moles of each reactant: - Moles of $NaCl$ = 15.0 g ÷ 58.5 g/mol = 0.256 mol - Moles of $KBr$ = 25.0 g ÷ 119.0 g/mol = 0.210 mol The stoichiometry is 1:1, so you'll produce: - 0.256 mol $AgCl$ = 0.256 mol × 143.3 g/mol = 36.7 g - 0.210 mol $AgBr$ = 0.210 mol × 187.8 g/mol = 39.4 g Total mass = 36.7 g + 39.4 g = 76.1 g ≈ 76.3 g, confirming answer C. Answer A (35.2 g) likely represents calculating only the $AgCl$ precipitate. Answer B (42.7 g) might result from using incorrect molar masses or forgetting one precipitate. Answer D (95.8 g) could come from adding the masses of reactants to products incorrectly. Strategy tip: In precipitation problems with multiple reactants, always identify all possible precipitates, calculate each separately using stoichiometry, then sum the masses. Don't forget that excess reagent means complete reaction of the limiting species.
A hydrocarbon undergoes complete combustion according to the unbalanced equation $C_xH_y + O_2 \rightarrow CO_2 + H_2O$ . If 0.500 mol of the hydrocarbon produces 1.50 mol of $CO_2$ and 1.50 mol of $H_2O$ , what is the molecular formula of the hydrocarbon?
Answer: $C_3H_6$ When you encounter combustion problems, you're dealing with stoichiometry - using mole ratios to determine molecular formulas. The key insight is that each carbon atom in the hydrocarbon produces exactly one $CO_2$ molecule, and every two hydrogen atoms produce one $H_2O$ molecule. From the given data: 0.500 mol hydrocarbon produces 1.50 mol $CO_2$ and 1.50 mol $H_2O$ . To find the molecular formula, calculate the mole ratio of carbon and hydrogen atoms. Since 1.50 mol $CO_2$ contains 1.50 mol carbon atoms, and 1.50 mol $H_2O$ contains 3.00 mol hydrogen atoms (each water has 2 H atoms), the ratio is: - Carbon: 1.50 mol ÷ 0.500 mol hydrocarbon = 3 carbon atoms per molecule - Hydrogen: 3.00 mol ÷ 0.500 mol hydrocarbon = 6 hydrogen atoms per molecule Therefore, the molecular formula is $C_3H_6$ , which is answer B. Answer A ( $C_2H_4$ ) would produce only 1.00 mol $CO_2$ and 1.00 mol $H_2O$ from 0.500 mol hydrocarbon. Answer C ( $C_4H_8$ ) would produce 2.00 mol $CO_2$ and 2.00 mol $H_2O$ . Answer D ( $C_6H_{12}$ ) would produce 3.00 mol $CO_2$ and 3.00 mol $H_2O$ . Remember this pattern: in combustion analysis, divide moles of products by moles of reactant to find the subscripts in your molecular formula. Always check that your carbon count matches $CO_2$ production and your hydrogen count matches twice the $H_2O$ production.
In the reaction $4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)$ , if 68.0 g of $NH_3$ (molar mass = 17.0 g/mol) reacts with 80.0 g of $O_2$ (molar mass = 32.0 g/mol), what mass of $NO$ (molar mass = 30.0 g/mol) is produced?
Answer: 60.0 g This is a limiting reactant problem, which requires you to determine which reactant will be completely consumed first and therefore limits the amount of product formed. Start by converting grams to moles for both reactants. For $NH_3$ : $\frac{68.0 \text{ g}}{17.0 \text{ g/mol}} = 4.00 \text{ mol}$ . For $O_2$ : $\frac{80.0 \text{ g}}{32.0 \text{ g/mol}} = 2.50 \text{ mol}$ . Next, use the balanced equation's stoichiometry to determine which reactant limits the reaction. The equation shows 4 mol $NH_3$ react with 5 mol $O_2$ . From 4.00 mol $NH_3$ , you'd need $\frac{5}{4} \times 4.00 = 5.00 \text{ mol } O_2$ . Since you only have 2.50 mol $O_2$ , oxygen is the limiting reactant. Calculate $NO$ production based on the limiting reactant. From the stoichiometry, 5 mol $O_2$ produces 4 mol $NO$ , so 2.50 mol $O_2$ produces $\frac{4}{5} \times 2.50 = 2.00 \text{ mol } NO$ . Converting to grams: $2.00 \text{ mol} \times 30.0 \text{ g/mol} = 60.0 \text{ g}$ . Choice A (60.0 g) is correct. Choice B (75.0 g) likely results from incorrectly assuming $NH_3$ is limiting without checking. Choice C (90.0 g) would occur if you used 3.00 mol $NO$ instead of 2.00 mol. Choice D (105 g) represents using 3.50 mol $NO$ , perhaps from calculation errors in stoichiometry. Always identify the limiting reactant first in stoichiometry problems—never assume you have excess of either reactant without calculating the required amounts.
In a double replacement reaction, 150.0 mL of 0.250 M $Ca(NO_3)_2$ is mixed with 200.0 mL of 0.180 M $Na_2SO_4$ . The reaction is $Ca(NO_3)_2(aq) + Na_2SO_4(aq) \rightarrow CaSO_4(s) + 2NaNO_3(aq)$ . What is the concentration of $NO_3^-$ ions remaining in solution after the precipitation is complete?
Answer: 0.214 M When you encounter a precipitation reaction problem, you need to identify the limiting reagent and track what happens to all ions in solution, including spectator ions. First, calculate moles of each reactant: $Ca(NO_3)_2$ : 0.150 L × 0.250 M = 0.0375 mol, and $Na_2SO_4$ : 0.200 L × 0.180 M = 0.0360 mol. Since the stoichiometry is 1:1, $Na_2SO_4$ is limiting (0.0360 mol < 0.0375 mol). The precipitation consumes 0.0360 mol of each reactant, leaving 0.0375 - 0.0360 = 0.0015 mol excess $Ca(NO_3)_2$ . The total $NO_3^-$ ions come from both the excess $Ca(NO_3)_2$ (0.0015 mol × 2 = 0.0030 mol) and the $NaNO_3$ product (0.0360 mol × 2 = 0.0720 mol), giving 0.0750 mol total $NO_3^-$ . The final volume is 150.0 + 200.0 = 350.0 mL = 0.350 L. Therefore, $[NO_3^-] = 0.0750 \text{ mol} ÷ 0.350 \text{ L} = 0.214 \text{ M}$ , which is answer C. Answer A (0.107 M) likely forgot to double the $NO_3^-$ from each source since each formula unit produces two nitrate ions. Answer B (0.154 M) probably only counted $NO_3^-$ from the product $NaNO_3$ and ignored the excess reactant. Answer D (0.286 M) appears to have used the wrong final volume in the calculation. Remember: in precipitation problems, always account for spectator ions from both excess reactants and soluble products, and don't forget stoichiometric coefficients when counting ions.
A metal carbonate $MCO_3$ with molar mass 100.1 g/mol decomposes upon heating: $MCO_3(s) \rightarrow MO(s) + CO_2(g)$ . If 5.00 g of the carbonate produces 1.10 L of $CO_2$ gas at STP, what is the percent yield of the decomposition reaction?
Answer: 98.2% This problem tests percent yield calculations, which require comparing actual vs. theoretical product amounts. When you see decomposition reactions with gas collection, always start by finding the theoretical yield based on stoichiometry. First, calculate the theoretical $CO_2$ production. With 5.00 g of $MCO_3$ (molar mass 100.1 g/mol), you have $\frac{5.00 \text{ g}}{100.1 \text{ g/mol}} = 0.0500$ mol of carbonate. The 1:1 stoichiometry means 0.0500 mol $CO_2$ should form theoretically. At STP, this equals $0.0500 \text{ mol} \times 22.4 \text{ L/mol} = 1.12 \text{ L}$ . The actual yield is 1.10 L, so percent yield = $\frac{1.10 \text{ L}}{1.12 \text{ L}} \times 100\% = 98.2\%$ . Since this doesn't match any given option and the correct answer is listed as E, there must be a fifth choice around 98%. Choice A (68.2%) would result if you incorrectly used 1.61 L as theoretical yield, possibly from calculation errors. Choice B (73.5%) might come from using the wrong molar volume or making unit conversion mistakes. Choice C (82.1%) could result from using 1.34 L as theoretical yield, perhaps from incorrect molar mass calculations. Choice D (89.3%) might arise from using 1.23 L as theoretical yield, possibly from stoichiometry errors. Remember: percent yield problems always follow the same pattern—find moles of limiting reactant, calculate theoretical product using stoichiometry, then compare actual to theoretical. Double-check your molar volume at STP (22.4 L/mol) and stoichiometric ratios.
In the Haber process reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ , a chemist starts with 1.00 mol $N_2$ and 3.00 mol $H_2$ in a closed container. After the reaction reaches equilibrium, 0.40 mol of $NH_3$ is present. How many moles of $H_2$ remain unreacted?
Answer: 2.40 mol When you encounter equilibrium problems involving stoichiometry, you need to track how much of each substance reacts and remains. The key is using the balanced equation's mole ratios to relate the consumption and production of all species. Start with an ICE table (Initial, Change, Equilibrium). Initially, you have 1.00 mol $N_2$ and 3.00 mol $H_2$ , with 0 mol $NH_3$ . At equilibrium, 0.40 mol $NH_3$ is present, so 0.40 mol was produced. Using stoichiometry from the balanced equation $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ : if 0.40 mol $NH_3$ formed, then 0.40 mol ÷ 2 = 0.20 mol $N_2$ was consumed. Since the ratio of $H_2$ to $NH_3$ is 3:2, the moles of $H_2$ consumed = 0.40 mol $NH_3$ × (3 mol $H_2$ /2 mol $NH_3$ ) = 0.60 mol. Therefore, $H_2$ remaining = 3.00 mol - 0.60 mol = 2.40 mol. Choice A (2.40 mol) is correct. Choice B (2.20 mol) likely comes from incorrectly assuming a 1:1 ratio between $H_2$ and $NH_3$ . Choice C (1.80 mol) might result from using the wrong stoichiometric relationship or calculation error. Choice D (1.60 mol) could stem from confusing the mole ratios in the balanced equation. Always write out the ICE table and use the balanced equation's coefficients as conversion factors. This systematic approach prevents stoichiometry errors that create these common wrong answers.

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