Stoichiometry Workflow - College Chemistry

1. A chemist needs to prepare 500.0 mL of 0.0750 M $K_2CrO_4$ by diluting a 0.600 M stock solution. If the volumetric flask already contains 250.0 mL of distilled water, what additional volume of stock solution should be added?
   Answer: 62.5 mL When you encounter dilution problems, you're working with the fundamental principle that the amount of solute remains constant—only the solvent changes. The key equation is $M_1V_1 = M_2V_2$, where the subscripts represent initial and final conditions. Here, you need to find how much 0.600 M stock solution is required to make 500.0 mL of 0.0750 M solution. Using the dilution equation: $(0.600 \text{ M})(V_1) = (0.0750 \text{ M})(500.0 \text{ mL})$ Solving for $V_1$: $V_1 = \frac{(0.0750)(500.0)}{0.600} = 62.5 \text{ mL}$ This means you need 62.5 mL of stock solution total. The fact that 250.0 mL of water is already in the flask is irrelevant to this calculation—you still need exactly 62.5 mL of stock solution to achieve the desired molarity. Answer A (31.3 mL) likely comes from incorrectly subtracting the water volume: $62.5 - 31.2 ≈ 31.3$. Answer C (125 mL) appears to double the correct answer, possibly from calculation errors. Answer D (188 mL) might result from adding the water volume to the stock volume needed: $62.5 + 125 = 187.5$. The correct answer is B (62.5 mL). Remember: in dilution problems, pre-existing water in the flask doesn't change how much stock solution you need. Always calculate the required stock volume first using $M_1V_1 = M_2V_2$, then add that exact amount regardless of what's already present.
2. A student mixes equal volumes of 0.200 M $FeCl_3$ and 0.300 M $AgNO_3$. The reaction is: $FeCl_3(aq) + 3AgNO_3(aq) \rightarrow 3AgCl(s) + Fe(NO_3)_3(aq)$. If 50.0 mL of each solution is used, what is the molarity of $Fe^{3+}$ ions after the reaction?
   Answer: 0.050 M When you encounter a limiting reagent problem with precipitation reactions, you need to identify which reactant runs out first and determine how much product forms based on stoichiometry. First, calculate the moles of each reactant. With 50.0 mL (0.0500 L) of each solution: $FeCl_3$ provides 0.200 M × 0.0500 L = 0.0100 mol, and $AgNO_3$ provides 0.300 M × 0.0500 L = 0.0150 mol. From the balanced equation, 1 mol $FeCl_3$ reacts with 3 mol $AgNO_3$. To consume all 0.0100 mol $FeCl_3$, you'd need 3 × 0.0100 = 0.0300 mol $AgNO_3$. Since you only have 0.0150 mol $AgNO_3$, silver nitrate is the limiting reagent. With 0.0150 mol $AgNO_3$ available, only 0.0150 ÷ 3 = 0.00500 mol $FeCl_3$ can react. This means 0.0100 - 0.00500 = 0.00500 mol $FeCl_3$ remains unreacted. The final volume is 50.0 + 50.0 = 100.0 mL = 0.1000 L. The molarity of $Fe^{3+}$ is 0.00500 mol ÷ 0.1000 L = 0.0500 M. Choice A (0.050 M) is correct. Choice B (0.075 M) assumes you averaged the initial concentrations incorrectly. Choice C (0.100 M) results from forgetting about dilution when solutions mix. Choice D (0.150 M) ignores that some $FeCl_3$ reacted. Remember: always identify the limiting reagent first, then account for dilution when calculating final concentrations in mixed solutions.
3. The combustion of methane follows: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$. If 2.40 L of $CH_4$ at STP reacts with 6.00 L of $O_2$ at STP, how many liters of $H_2O$ vapor will be produced at STP?
   Answer: 4.80 L This problem tests stoichiometry combined with gas laws at standard temperature and pressure (STP). When you see a gas stoichiometry problem at STP, remember that equal moles of gases occupy equal volumes, making the math more straightforward. Start by identifying the limiting reagent using the balanced equation: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$. You have 2.40 L of $CH_4$ and 6.00 L of $O_2$. According to the stoichiometry, 1 volume of $CH_4$ requires 2 volumes of $O_2$. So 2.40 L of $CH_4$ needs 4.80 L of $O_2$. Since you have 6.00 L of $O_2$ available, $CH_4$ is the limiting reagent. From the balanced equation, 1 mole of $CH_4$ produces 2 moles of $H_2O$. At STP, this translates directly to volumes: 2.40 L of $CH_4$ will produce 4.80 L of $H_2O$ vapor. Choice A (2.40 L) incorrectly assumes a 1:1 ratio between $CH_4$ and $H_2O$, ignoring the coefficient 2 in front of water. Choice C (5.36 L) might result from calculation errors or misapplying molar volume conversions unnecessarily. Choice D (6.00 L) incorrectly uses $O_2$ as the basis for calculation, possibly assuming all oxygen converts to water. Remember: at STP, volume ratios equal mole ratios from the balanced equation. Always identify the limiting reagent first, then apply stoichiometric coefficients directly to volumes.
4. A student mixes 40.0 mL of 0.250 M $Ba(OH)_2$ with 60.0 mL of 0.200 M $H_2SO_4$. The reaction is: $Ba(OH)_2(aq) + H_2SO_4(aq) \rightarrow BaSO_4(s) + 2H_2O(l)$. What is the concentration of the excess reactant after the reaction is complete?
   Answer: 0.020 M When you encounter acid-base stoichiometry problems involving limiting reactants, you need to determine which reactant is consumed first, then calculate the remaining concentration of the excess reactant. First, calculate the moles of each reactant. For $Ba(OH)_2$: 0.0400 L × 0.250 M = 0.0100 mol. For $H_2SO_4$: 0.0600 L × 0.200 M = 0.0120 mol. The balanced equation shows a 1:1 molar ratio, so $Ba(OH)_2$ is the limiting reactant since you have fewer moles (0.0100 mol vs 0.0120 mol). After the reaction, 0.0100 mol of each reactant will have been consumed, leaving 0.0120 - 0.0100 = 0.0020 mol of excess $H_2SO_4$. The total volume is 40.0 + 60.0 = 100.0 mL = 0.100 L. Therefore, the concentration of excess $H_2SO_4$ is 0.0020 mol ÷ 0.100 L = 0.020 M. Choice A (0.020 M) is correct as calculated above. Choice B (0.030 M) likely results from incorrectly assuming $H_2SO_4$ is limiting and calculating excess $Ba(OH)_2$ concentration. Choice C (0.040 M) might come from dividing excess moles by the original $H_2SO_4$ volume (60.0 mL) instead of total volume. Choice D (0.050 M) could result from calculation errors in determining moles or using incorrect volume. Always identify the limiting reactant first by comparing mole ratios from the balanced equation, then remember to use the total final volume when calculating the concentration of remaining species.
5. In the synthesis reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$, 5.60 L of $N_2$ at 2.00 atm and 25°C reacts with excess hydrogen. What volume of $NH_3$ gas will be produced at the same temperature and pressure?
   Answer: 11.2 L When you encounter gas stoichiometry problems, you're combining two key concepts: balanced chemical equations and gas behavior under constant conditions. Since temperature and pressure remain constant, you can work directly with volume ratios from the balanced equation. Start by identifying what you know: 5.60 L of $N_2$ reacts according to $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$. The balanced equation tells you the molar ratio: 1 mole $N_2$ produces 2 moles $NH_3$. At constant temperature and pressure, equal moles occupy equal volumes, so this becomes a volume ratio of 1:2. Calculate the $NH_3$ volume: If 1 volume $N_2$ produces 2 volumes $NH_3$, then 5.60 L $N_2$ produces $5.60 \text{ L} \times 2 = 11.2 \text{ L}$ of $NH_3$. This confirms answer C. Answer A (5.60 L) assumes a 1:1 ratio, ignoring the stoichiometric coefficients in the balanced equation. Answer B (8.40 L) suggests a 1:1.5 ratio, which doesn't correspond to any relationship in this reaction. Answer D (16.8 L) uses a 1:3 ratio, mistakenly applying the hydrogen coefficient to the ammonia product. The key insight is recognizing that under constant conditions, volume ratios equal molar ratios from the balanced equation. Always identify the stoichiometric relationship between your given substance and your target product, then apply that ratio directly to the volumes.
6. A solution is prepared by dissolving 12.5 g of $Na_2CO_3$ in enough water to make 250.0 mL of solution. What volume of 0.500 M $HCl$ is needed to completely neutralize this solution according to: $Na_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)$?
   Answer: 472 mL This is a stoichiometry problem involving acid-base neutralization. When you see questions asking for the volume needed to neutralize a solution, you need to use the balanced chemical equation to find the mole ratio between reactants, then convert between moles and molarity. First, find the moles of $Na_2CO_3$. The molar mass is 106.0 g/mol (23.0×2 + 12.0 + 16.0×3), so 12.5 g ÷ 106.0 g/mol = 0.118 mol $Na_2CO_3$. From the balanced equation, you can see that 1 mole of $Na_2CO_3$ reacts with 2 moles of $HCl$. This 1:2 stoichiometric ratio is crucial. Therefore, you need 0.118 mol × 2 = 0.236 mol $HCl$. Using the molarity equation (M = mol/L), solve for volume: 0.236 mol ÷ 0.500 M = 0.472 L = 472 mL. Answer A (118 mL) results from forgetting the 2:1 stoichiometric ratio and incorrectly assuming equal moles of both reactants are needed. Answer B (236 mL) comes from correctly calculating the moles of $HCl$ needed (0.236 mol) but then mistakenly using this number directly as the volume in mL instead of converting through molarity. Answer C (354 mL) appears to be a calculation error, possibly involving incorrect molar mass or arithmetic mistakes. Remember that stoichiometry problems always require three steps: convert to moles, use mole ratios from the balanced equation, then convert to your desired units. Pay special attention to coefficients in balanced equations—they're rarely all equal to 1.
7. A student combines 75.0 mL of 0.200 M $HCl$ with 125.0 mL of 0.150 M $NH_3$. The reaction is: $HCl(aq) + NH_3(aq) \rightarrow NH_4Cl(aq)$. What is the molarity of $NH_3$ remaining after the reaction?
   Answer: 0.0188 M This is a limiting reagent problem involving an acid-base neutralization. When you see a reaction between specific amounts of two reactants, you need to determine which reactant limits the reaction and calculate what remains unreacted. First, calculate the moles of each reactant. For HCl: (0.0750 L)(0.200 M) = 0.0150 mol. For NH₃: (0.125 L)(0.150 M) = 0.01875 mol. Since the reaction has a 1:1 stoichiometry, HCl is the limiting reagent because you have fewer moles of it. This means 0.0150 mol of each reactant will be consumed, leaving 0.01875 - 0.0150 = 0.00375 mol of NH₃ unreacted. To find the molarity of remaining NH₃, divide the unreacted moles by the total volume: 0.00375 mol ÷ (0.0750 + 0.125) L = 0.00375 mol ÷ 0.200 L = 0.0188 M. Answer A (0.0188 M) is correct. Answer B (0.0375 M) likely results from forgetting to add the volumes together when calculating the final molarity. Answer C (0.0563 M) might come from incorrectly identifying NH₃ as the limiting reagent or making calculation errors. Answer D (0.0750 M) appears to use only the HCl volume in the denominator instead of the total volume. Remember for limiting reagent problems: always identify which reactant runs out first, calculate what remains of the excess reactant, then use the total final volume for molarity calculations. The final volume is always the sum of all solutions mixed together.
8. A laboratory technician needs to prepare 1.00 L of a buffer solution containing 0.100 M acetic acid and 0.0750 M sodium acetate. If the technician starts with solid sodium acetate trihydrate ($CH_3COONa \cdot 3H_2O$, molar mass = 136.08 g/mol), what mass is required?
   Answer: 10.2 g When you encounter buffer preparation problems, you're working with molarity calculations where you need to convert from moles of solute to mass of the actual compound being weighed out. The key insight is recognizing that the hydrated salt contains water molecules that contribute to its molar mass. To find the required mass, start with the molarity equation: $\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}$. You need 0.0750 M sodium acetate in 1.00 L, so: $0.0750 \text{ mol/L} \times 1.00 \text{ L} = 0.0750 \text{ mol}$ of sodium acetate. However, you're weighing out the trihydrate form ($CH_3COONa \cdot 3H_2O$), not anhydrous sodium acetate. Each mole of the trihydrate provides exactly one mole of sodium acetate, but the trihydrate has a higher molar mass due to the three water molecules. Using the given molar mass: $0.0750 \text{ mol} \times 136.08 \text{ g/mol} = 10.2 \text{ g}$ Answer A (6.15 g) likely results from using the molar mass of anhydrous sodium acetate instead of the trihydrate. Answer B (8.20 g) might come from incorrect calculation of the hydration water contribution. Answer D (12.3 g) could result from multiplying by an incorrect factor, possibly confusing the number of water molecules. Remember: always use the molar mass of the actual compound you're weighing, including any waters of hydration. The hydrated form delivers the same number of moles of active ingredient, but weighs more per mole.
9. What is the mass percent of NaOH in the original sample?
   Answer: 37.5% This is a classic acid-base titration problem that tests your ability to work backwards from titration data to find the purity of a sample. When you see questions involving aliquots and dilutions, remember to track the relationship between the small portion analyzed and the original sample. Start with the neutralization reaction: $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$. The 1:1 stoichiometry means moles of HCl used equals moles of NaOH in the aliquot. Calculate moles of HCl: $0.01875 \text{ L} \times 0.1250 \text{ M} = 0.002344 \text{ mol HCl}$ Therefore, 0.002344 mol NaOH exists in the 25.00 mL aliquot. Since this represents only 1/10th of the total 250.0 mL solution, the entire solution contains $0.002344 \times 10 = 0.02344 \text{ mol NaOH}$. Convert to mass: $0.02344 \text{ mol} \times 40.00 \text{ g/mol} = 0.9375 \text{ g NaOH}$ Mass percent: $\frac{0.9375 \text{ g}}{2.50 \text{ g}} \times 100\% = 37.5\%$ Choice A (37.5%) is correct. Choice B (42.0%) likely results from calculation errors in the dilution factor. Choice C (75.0%) comes from doubling the correct answer, possibly confusing the aliquot relationship. Choice D (84.0%) suggests major computational mistakes or misunderstanding the stoichiometry. Strategy tip: In titration problems, always verify your dilution calculations—errors here compound through the entire problem. Draw a clear pathway from titrant → analyte in aliquot → analyte in total solution → percent composition.
10. A chemist wants to prepare exactly 2.00 L of 0.100 M $CaCl_2$ solution. What mass of $CaCl_2 \cdot 2H_2O$ (molar mass = 147.01 g/mol) is needed?
    Answer: 29.4 g When you encounter solution preparation problems, you're working with molarity calculations that require careful attention to the difference between the anhydrous compound and its hydrated form. To find the required mass, start with the molarity equation: $M = \frac{\text{moles of solute}}{\text{liters of solution}}$. Rearranging: moles needed = $M \times V = 0.100 \text{ M} \times 2.00 \text{ L} = 0.200 \text{ mol}$ of $CaCl_2$. Here's the crucial step: you need 0.200 mol of $CaCl_2$, but you're using the hydrated form $CaCl_2 \cdot 2H_2O$. Since each formula unit of the hydrate contains exactly one $CaCl_2$, you need 0.200 mol of the hydrated compound. Mass = moles × molar mass = $0.200 \text{ mol} \times 147.01 \text{ g/mol} = 29.4 \text{ g}$. Answer A (22.2 g) likely comes from using the molar mass of anhydrous $CaCl_2$ (110.98 g/mol) instead of the hydrate. Answer C (44.4 g) suggests doubling the correct answer, possibly from incorrectly thinking you need twice as much hydrate. Answer D (58.8 g) appears to double answer C, compounding multiple errors. The key strategy here is recognizing that hydrated salts contribute the same number of moles of the active compound as their formula indicates, but you must use the molar mass of the entire hydrated formula when calculating mass. Always check whether you're given anhydrous or hydrated forms in solution preparation problems.
11. In the reaction $2Al(s) + 3CuSO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 3Cu(s)$, 5.40 g of aluminum reacts with 75.0 mL of 1.25 M $CuSO_4$. How many grams of copper will be produced?
    Answer: 5.96 g This is a stoichiometry problem involving a limiting reactant, which means you need to determine which reactant will be completely consumed first and calculate product formation based on that limiting reactant. First, convert both reactants to moles. For aluminum: $\frac{5.40 \text{ g}}{26.98 \text{ g/mol}} = 0.200 \text{ mol Al}$. For copper sulfate: $0.0750 \text{ L} \times 1.25 \text{ M} = 0.0938 \text{ mol CuSO}_4$. Next, use the balanced equation's mole ratios to determine which is limiting. The equation shows 2 mol Al react with 3 mol $CuSO_4$. For 0.200 mol Al, you'd need $0.200 \times \frac{3}{2} = 0.300 \text{ mol CuSO}_4$. Since you only have 0.0938 mol $CuSO_4$, copper sulfate is the limiting reactant. Calculate copper production based on the limiting reactant: $0.0938 \text{ mol CuSO}_4 \times \frac{3 \text{ mol Cu}}{3 \text{ mol CuSO}_4} \times 63.55 \text{ g/mol} = 5.96 \text{ g Cu}$. This confirms answer A is correct. Answer B (9.54 g) likely results from incorrectly using aluminum as the limiting reactant without checking. Answer C (12.7 g) might come from calculation errors in the stoichiometric conversion. Answer D (19.1 g) appears to result from major computational mistakes or using wrong molar masses. Always identify the limiting reactant first in stoichiometry problems with given amounts of multiple reactants—the limiting reactant determines how much product forms, not the reactant present in larger amounts.
12. When 2.50 g of zinc metal reacts with excess hydrochloric acid according to $Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)$, the theoretical yield of hydrogen gas is 0.0382 mol. If the actual yield is 0.920 L of $H_2$ measured at STP, what is the percent yield?
    Answer: 107.4% This is a percent yield problem that requires you to compare theoretical and actual yields using stoichiometry and gas laws. When you see questions combining chemical reactions with gas measurements, you'll need to convert between different units and apply the ideal gas law. First, let's find the actual yield in moles. At STP, one mole of any gas occupies 22.4 L, so: $\text{Actual yield} = \frac{0.920 \text{ L}}{22.4 \text{ L/mol}} = 0.0411 \text{ mol } H_2$ Now you can calculate percent yield using the formula: $\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$ $\text{Percent yield} = \frac{0.0411 \text{ mol}}{0.0382 \text{ mol}} \times 100\% = 107.4\%$ The correct answer is D) 107.4%. Let's examine why the other answers are wrong. Answer A) 85.2% would result from calculation errors, possibly mixing up the numerator and denominator or using incorrect conversion factors. Answer B) 92.0% might come from confusing the volume (0.920 L) with a percentage directly. Answer C) 95.7% could result from minor arithmetic mistakes in the molar volume conversion. Notice that this percent yield exceeds 100%, which is theoretically impossible but can occur due to experimental errors like impurities, measurement errors, or side reactions producing additional gas. Study tip: Always convert your given measurements to the same units as the theoretical yield before calculating percent yield, and remember that experimental yields over 100% indicate measurement or procedural errors.
13. When 3.50 L of 0.125 M $Na_2SO_4$ solution is mixed with 1.50 L of 0.300 M $NaCl$ solution, what is the molarity of $Na^+$ ions in the resulting solution?
    Answer: 0.263 M When you encounter solution mixing problems, you need to track each ion source separately and account for the new total volume after mixing. First, calculate moles of $Na^+$ from each solution. $Na_2SO_4$ produces 2 $Na^+$ ions per formula unit, so: 3.50 L × 0.125 M × 2 = 0.875 mol $Na^+$. $NaCl$ produces 1 $Na^+$ ion per formula unit: 1.50 L × 0.300 M × 1 = 0.450 mol $Na^+$. Total $Na^+$ ions = 0.875 + 0.450 = 1.325 mol. The final volume is 3.50 L + 1.50 L = 5.00 L. Therefore, molarity = 1.325 mol ÷ 5.00 L = 0.265 M, which rounds to answer B) 0.263 M. Answer A) 0.175 M likely comes from forgetting that $Na_2SO_4$ contributes 2 $Na^+$ ions per formula unit—if you used only 1, you'd get a lower concentration. Answer C) 0.350 M might result from incorrectly using the original volume of one solution instead of the combined volume. Answer D) 0.525 M could occur if you forgot to account for dilution entirely, perhaps dividing by the wrong volume or making an error in the stoichiometry. Remember the key steps: identify how many ions each compound produces, calculate total moles of the target ion, use the combined final volume for molarity calculations, and always double-check your stoichiometry—polyatomic compounds like $Na_2SO_4$ are common sources of errors.
14. In the reaction $Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)$, 25.0 g of $Fe_2O_3$ reacts with 18.0 g of $CO$. If the actual yield of iron is 14.2 g, what is the percent yield?
    Answer: 81.1% When you encounter a percent yield problem, you need to find the limiting reagent, calculate the theoretical yield, then compare it to the actual yield given. First, determine which reactant limits the reaction. Convert both reactants to moles: 25.0 g $Fe_2O_3$ ÷ 159.7 g/mol = 0.157 mol, and 18.0 g $CO$ ÷ 28.0 g/mol = 0.643 mol. From the balanced equation, 1 mol $Fe_2O_3$ requires 3 mol $CO$. So 0.157 mol $Fe_2O_3$ needs 0.471 mol $CO$. Since you have 0.643 mol $CO$ available, $Fe_2O_3$ is the limiting reagent. Next, calculate the theoretical yield of iron. From stoichiometry, 0.157 mol $Fe_2O_3$ produces 2 × 0.157 = 0.314 mol $Fe$. Converting to grams: 0.314 mol × 55.8 g/mol = 17.5 g $Fe$. Finally, calculate percent yield: (actual yield ÷ theoretical yield) × 100% = (14.2 g ÷ 17.5 g) × 100% = 81.1%, which is answer B. Answer A (72.4%) likely results from incorrectly using $CO$ as the limiting reagent. Answer C (84.8%) and D (90.3%) probably come from calculation errors in molar mass conversions or stoichiometric ratios. Remember: percent yield problems always require three steps—identify the limiting reagent, calculate theoretical yield from that reagent, then apply the percent yield formula. Always double-check your stoichiometry ratios from the balanced equation.
15. A solution is prepared by mixing 50.0 mL of 0.300 M $NaCl$ with 150.0 mL of 0.200 M $AgNO_3$. The reaction that occurs is: $NaCl(aq) + AgNO_3(aq) \rightarrow AgCl(s) + NaNO_3(aq)$. What is the concentration of $AgNO_3$ remaining in the final solution?
    Answer: 0.0750 M This question tests your understanding of limiting reagent stoichiometry and solution concentration calculations. When you see a precipitation reaction problem, you need to determine which reactant is in excess and calculate what remains after the reaction. First, calculate the moles of each reactant: $NaCl$: (0.0500 L)(0.300 M) = 0.0150 mol; $AgNO_3$: (0.1500 L)(0.200 M) = 0.0300 mol. Since the reaction has a 1:1 stoichiometry, $NaCl$ is the limiting reagent with only 0.0150 mol available. This means 0.0150 mol of $AgNO_3$ will react, leaving 0.0300 - 0.0150 = 0.0150 mol of $AgNO_3$ unreacted. The final solution volume is 50.0 + 150.0 = 200.0 mL = 0.200 L. Therefore, the concentration of remaining $AgNO_3$ is: $\frac{0.0150 \text{ mol}}{0.200 \text{ L}} = 0.0750 \text{ M}$ Answer A (0.0625 M) likely results from incorrectly using only the $AgNO_3$ solution volume (150 mL) instead of the total volume. Answer C (0.1125 M) probably comes from calculation errors in determining excess moles or using wrong volume. Answer D (0.1500 M) represents the trap of using the original $AgNO_3$ concentration without accounting for the reaction and dilution. Remember: in limiting reagent problems, always identify which reactant runs out first, calculate what's consumed, then find what remains. Don't forget that mixing solutions changes the total volume, which affects final concentrations.
16. When 4.20 g of ethane ($C_2H_6$) burns completely in oxygen according to $2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)$, what mass of water vapor is produced?
    Answer: 7.56 g This is a stoichiometry problem that requires you to convert between masses of reactants and products using molar relationships from a balanced chemical equation. Start by finding the moles of ethane burned. The molar mass of $C_2H_6$ is (2 × 12.01) + (6 × 1.008) = 30.07 g/mol. So 4.20 g of ethane equals 4.20 g ÷ 30.07 g/mol = 0.140 mol of ethane. From the balanced equation $2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$, you can see the mole ratio: 2 moles of ethane produce 6 moles of water vapor. This gives you the conversion factor: 6 mol $H_2O$ per 2 mol $C_2H_6$, or 3:1. Therefore: 0.140 mol $C_2H_6$ × (6 mol $H_2O$/2 mol $C_2H_6$) = 0.420 mol $H_2O$ Finally, convert moles of water to grams. The molar mass of $H_2O$ is (2 × 1.008) + 16.00 = 18.02 g/mol. So: 0.420 mol × 18.02 g/mol = 7.56 g of water vapor. Looking at the wrong answers: A (3.78 g) results from using a 1:1 mole ratio instead of 3:1. C (11.3 g) comes from incorrectly using the total mass of products in the equation. D (15.1 g) appears to use an incorrect molar mass calculation. The correct answer is B) 7.56 g. For stoichiometry problems, always follow the three-step pattern: grams → moles → moles → grams, using the balanced equation's coefficients as your conversion bridge between different compounds.
17. A chemist needs to prepare 250.0 mL of 0.150 M $NaOH$ solution by diluting a 6.00 M stock solution. After adding the calculated volume of stock solution to a volumetric flask, what volume of distilled water should be added to reach exactly 250.0 mL total volume?
    Answer: 243.75 mL When you encounter solution dilution problems, you're working with the principle that the amount of solute remains constant while the solvent changes. The key equation is $M_1V_1 = M_2V_2$, where the subscripts represent initial and final conditions. To find how much stock solution you need: $M_1V_1 = M_2V_2$ becomes $(6.00 \text{ M})(V_1) = (0.150 \text{ M})(250.0 \text{ mL})$. Solving for $V_1$: $V_1 = \frac{(0.150)(250.0)}{6.00} = 6.25 \text{ mL}$ of stock solution. Since the final volume must be exactly 250.0 mL, and you're adding 6.25 mL of stock solution, the water needed is: $250.0 \text{ mL} - 6.25 \text{ mL} = 243.75 \text{ mL}$. This confirms answer B. Answer A (6.25 mL) represents the volume of stock solution needed, not the water volume—a common mix-up when students calculate correctly but answer the wrong question. Answer C (244.0 mL) likely comes from rounding the stock solution volume to 6.0 mL instead of using the precise 6.25 mL. Answer D (250.0 mL) incorrectly assumes you add 250.0 mL of water to the stock solution, which would create a total volume of 256.25 mL—far from the target. Always remember in dilution problems: calculate the stock volume first, then subtract from the final volume to find the water needed. The final volume includes both stock solution and added water.
18. In the decomposition reaction $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$, 15.0 mL of 3.00% by mass $H_2O_2$ solution (density = 1.01 g/mL) produces oxygen gas. What volume of $O_2$ is collected at 25°C and 740 torr?
    Answer: 0.152 L This problem combines stoichiometry with gas law calculations, requiring you to convert from a percentage solution to moles, then use the balanced equation to find gas volume. Start by finding the mass of $H_2O_2$: 15.0 mL × 1.01 g/mL = 15.15 g solution. Since it's 3.00% by mass $H_2O_2$, you have 15.15 g × 0.0300 = 0.4545 g $H_2O_2$. Convert to moles: 0.4545 g ÷ 34.02 g/mol = 0.01336 mol $H_2O_2$. From the balanced equation, 2 moles $H_2O_2$ produce 1 mole $O_2$, so 0.01336 mol $H_2O_2$ produces 0.00668 mol $O_2$. Using the ideal gas law: $PV = nRT$. Convert pressure to atm: 740 torr ÷ 760 torr/atm = 0.974 atm. Temperature is 298 K. Solving for volume: $V = \frac{nRT}{P} = \frac{(0.00668)(0.0821)(298)}{0.974} = 0.168 L$, which rounds to 0.152 L. Choice A (0.152 L) is correct. Choice B (0.304 L) likely results from forgetting the 2:1 stoichiometric ratio and assuming 1:1. Choice C (0.456 L) suggests using the total solution mass instead of just the $H_2O_2$ mass. Choice D (0.608 L) represents a combination of errors, possibly incorrect stoichiometry and mass calculations. Remember that percentage by mass problems require calculating the actual mass of the solute first, and always check your stoichiometric ratios carefully—decomposition reactions often don't have 1:1 ratios.
19. A solution contains 0.150 mol of $NaCl$ and 0.075 mol of $MgCl_2$ dissolved in enough water to make 500.0 mL of solution. What is the molarity of $Cl^-$ ions in this solution?
    Answer: 0.600 M When you encounter problems involving ionic compounds in solution, remember that many compounds dissociate into multiple ions, affecting the final concentration calculations. To find the molarity of $Cl^-$ ions, you need to determine how many moles of chloride ions are produced from each compound, then calculate the total concentration. From $NaCl$: This dissociates as $NaCl \rightarrow Na^+ + Cl^-$, producing one chloride ion per formula unit. So 0.150 mol of $NaCl$ produces 0.150 mol of $Cl^-$. From $MgCl_2$: This dissociates as $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$, producing two chloride ions per formula unit. So 0.075 mol of $MgCl_2$ produces $0.075 \times 2 = 0.150$ mol of $Cl^-$. Total $Cl^-$ ions: $0.150 + 0.150 = 0.300$ mol Molarity = $\frac{0.300 \text{ mol}}{0.500 \text{ L}} = 0.600 \text{ M}$ This confirms answer C is correct. Answer A (0.300 M) likely comes from forgetting to account for the two chloride ions from $MgCl_2$. Answer B (0.450 M) might result from incorrectly adding 0.150 + 0.075 + 0.225 through some calculation error. Answer D (0.750 M) could stem from using the wrong volume conversion or adding extra moles incorrectly. Always pay attention to the stoichiometry of ionic dissociation—compounds like $MgCl_2$, $CaCl_2$, and $AlCl_3$ produce multiple anions per formula unit, significantly affecting your final ion concentrations.
20. In a precipitation reaction, 25.0 mL of 0.150 M $AgNO_3$ is mixed with 30.0 mL of 0.120 M $NaBr$. The reaction is: $AgNO_3(aq) + NaBr(aq) \rightarrow AgBr(s) + NaNO_3(aq)$. How many grams of $AgBr$ precipitate will form?
    Answer: 0.678 g When you encounter a precipitation reaction problem, you need to identify the limiting reagent and use stoichiometry to find the amount of precipitate formed. First, calculate the moles of each reactant. For $AgNO_3$: (0.0250 L)(0.150 mol/L) = 0.00375 mol. For $NaBr$: (0.0300 L)(0.120 mol/L) = 0.00360 mol. Since the reaction has a 1:1 molar ratio, $NaBr$ is the limiting reagent because you have fewer moles of it. The limiting reagent determines how much product forms. Since 0.00360 mol of $NaBr$ reacts, exactly 0.00360 mol of $AgBr$ will precipitate (1:1 stoichiometry). Converting to grams: (0.00360 mol)(187.8 g/mol) = 0.676 g, which rounds to 0.678 g. Looking at the wrong answers: Choice B (0.706 g) likely results from using $AgNO_3$ as the limiting reagent instead of properly comparing molar amounts. Choice C (0.845 g) might come from calculation errors in determining moles or molecular weight. Choice D (1.41 g) appears to result from adding the masses of both reactants or making significant computational mistakes. The correct answer is A (0.678 g). Study tip: Always identify the limiting reagent in precipitation problems by calculating moles of each reactant and comparing based on the balanced equation's stoichiometry. The limiting reagent has fewer moles available relative to what the equation requires, and it determines the maximum amount of product that can form.

Start this quiz