Pre AP Chemistry -3.2D Theoretical and Percent Yield- MCQ Exam Style Questions -New Syllabus 2025-2026
Pre AP Chemistry -Link- MCQ Exam Style Questions – New Syllabus 2025-2026
Pre AP Chemistry -Link- MCQ Exam Style Questions – Pre AP Chemistry – per latest Pre AP Chemistry Syllabus.
Calcium carbonate is heated. Calcium oxide and carbon dioxide gas are formed.
The equation for the reaction is shown.
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]
225 kg of calcium carbonate is heated until there is no further change in mass.
The yield of calcium oxide is 85 kg.
What is the percentage yield?
A) 37.8%
B) 47.2%
C) 67.5%
D) 85.0%
▶️ Answer/Explanation
Ans: C
First calculate the theoretical yield:
1. Molar mass of CaCO3 = 40 + 12 + (16×3) = 100 g/mol
2. Moles of CaCO3 = 225,000 g / 100 g/mol = 2,250 mol
3. From the equation, 1 mol CaCO3 produces 1 mol CaO
4. Molar mass of CaO = 40 + 16 = 56 g/mol
5. Theoretical yield = 2,250 mol × 56 g/mol = 126,000 g = 126 kg
Now calculate percentage yield:
Percentage yield = (actual yield / theoretical yield) × 100
= (85 kg / 126 kg) × 100 ≈ 67.5%
The equation for the reaction of iron(III) oxide with carbon monoxide is shown.
\[ Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2 \]
What is the percentage yield of iron when 16.8 g of carbon monoxide reacts completely with iron(III) oxide to form 8.96 g of iron?
A. 26.7%
B. 40.0%
C. 53.3%
D. 80.0%
▶️ Answer/Explanation
Ans: B
1. Calculate moles of CO:
Moles of CO = \(\frac{16.8 \text{ g}}{28 \text{ g/mol}} = 0.6 \text{ mol}\).
2. Determine theoretical yield of Fe:
From the equation, 3 moles of CO produce 2 moles of Fe.
Moles of Fe = \(\frac{2}{3} \times 0.6 = 0.4 \text{ mol}\).
Mass of Fe (theoretical) = \(0.4 \text{ mol} \times 56 \text{ g/mol} = 22.4 \text{ g}\).
3. Calculate percentage yield:
Percentage yield = \(\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{8.96 \text{ g}}{22.4 \text{ g}} \times 100 = 40.0\%\).
Thus, the correct answer is B (40.0%).
A tablet contains 0.080 g of ascorbic acid (\(M_{r}\) = 176). What is the concentration of ascorbic acid when one tablet is dissolved in 200 cm3 of water?
A. 9.1 × \(10^{-5}\) mol / \(dm^{3}\)
B. 4.5 × \(10^{-4}\) mol / \(dm^{3}\)
C. 9.1 × \(10^{-2}\) mol / \(dm^{3}\)
D. 2.3 × \(10^{-3}\) mol / \(dm^{3}\)
▶️ Answer/Explanation
Ans: D
1. Calculate moles of ascorbic acid:
\( \text{Moles} = \frac{\text{Mass}}{M_r} = \frac{0.080\, \text{g}}{176\, \text{g/mol}} = 4.55 \times 10^{-4}\, \text{mol} \)
2. Convert volume to dm3:
\( 200\, \text{cm}^3 = 0.200\, \text{dm}^3 \)
3. Calculate concentration:
\( \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} = \frac{4.55 \times 10^{-4}\, \text{mol}}{0.200\, \text{dm}^3} = 2.28 \times 10^{-3}\, \text{mol/dm}^3 \)
4. Round to match options:
The closest option is D (\( 2.3 \times 10^{-3}\, \text{mol/dm}^3 \)).
The Haber process is a reversible reaction.
\[ \mathrm{N}_2(\mathrm{~g}) + 3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) \]
The reaction has a 30% yield of ammonia.
Which volume of ammonia gas, \(\mathrm{NH}_3\), measured at room temperature and pressure, is obtained by reacting 0.75 moles of hydrogen with excess nitrogen?
A) \(3600 \mathrm{~cm}^3\)
B) \(5400 \mathrm{~cm}^3\)
C) \(12000 \mathrm{~cm}^3\)
D) \(18000 \mathrm{~cm}^3\)
▶️ Answer/Explanation
Ans: A
To determine the volume of ammonia produced:
- Stoichiometric ratio:
- The balanced equation shows that 3 moles of \(\mathrm{H}_2\) produce 2 moles of \(\mathrm{NH}_3\).
- Thus, 0.75 moles of \(\mathrm{H}_2\) would theoretically produce: \[ \frac{2}{3} \times 0.75 = 0.5 \text{ moles of } \mathrm{NH}_3. \]
- Account for 30% yield:
- Actual yield = \(0.5 \times 0.30 = 0.15\) moles of \(\mathrm{NH}_3\).
- Calculate volume at RTP:
- 1 mole of gas occupies \(24,000 \mathrm{~cm}^3\) at RTP.
- Volume of \(\mathrm{NH}_3\) = \(0.15 \times 24,000 = 3600 \mathrm{~cm}^3\).
Therefore, the correct answer is A (\(3600 \mathrm{~cm}^3\)).
Copper(II) carbonate is broken down by heating to form copper(II) oxide and carbon dioxide gas.
The equation for the reaction is shown.
CuCO3 → CuO + CO2
31.0g of copper(II) carbonate are heated until all of the contents of the test-tube have turned from green to black.
The yield of copper(II) oxide formed is 17.5g.
What is the percentage yield?
A) 19.02%
B) 21.88%
C) 56.50%
D) 87.50%
▶️ Answer/Explanation
Ans: D
Step 1: Calculate the theoretical yield of CuO
Molar mass of CuCO3 = 63.5 + 12 + (3 × 16) = 123.5 g/mol
Moles of CuCO3 = 31.0 g / 123.5 g/mol = 0.251 mol
From the equation: 1 mol CuCO3 produces 1 mol CuO
∴ Theoretical moles of CuO = 0.251 mol
Molar mass of CuO = 63.5 + 16 = 79.5 g/mol
Theoretical mass of CuO = 0.251 mol × 79.5 g/mol = 19.96 g
Step 2: Calculate percentage yield
Actual yield = 17.5 g
Percentage yield = (Actual yield / Theoretical yield) × 100
= (17.5 g / 19.96 g) × 100 ≈ 87.68%
The closest option is D (87.50%).
A student performs a reaction and collects 4.25 g of product. The theoretical yield of the product is 6.80 g.
Which of the following correctly calculates the percent yield, and which statement best explains why the percent yield is less than 100%?
A) 37.5% — because the reactants were not pure
B) 62.5% — because side reactions, incomplete reaction, or product loss during collection reduced the actual yield
C) 62.5% — because the theoretical yield calculation was incorrect
D) 160% — because more product than expected was formed
▶️ Answer/Explanation
Ans: B
Step 1: Apply the percent yield formula
$\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{4.25\,\text{g}}{6.80\,\text{g}} \times 100 = 62.5\%$
Step 2: Evaluate why percent yield < 100%
The theoretical yield is the maximum amount of product calculated from stoichiometry, assuming the reaction goes to completion with no losses. In practice, yield is reduced by:
• Side reactions — reactants form unwanted by-products instead of the desired product.
• Incomplete reaction — equilibrium or slow kinetics prevent full conversion of reactants.
• Product loss — product remains dissolved in solution, sticks to glassware, or is lost during filtration/transfer.
Option C is incorrect because a properly performed theoretical yield calculation using stoichiometry is assumed to be correct. The issue is experimental, not mathematical.
Therefore, the correct answer is B — 62.5%.
Magnesium burns in oxygen to form magnesium oxide:
2Mg + O2 → 2MgO
A student burns 6.00 g of magnesium (Mr = 24.3) in excess oxygen and collects 8.85 g of MgO (Mr = 40.3).
What is the theoretical yield of MgO, and what is the student’s percent yield?
A) Theoretical yield = 8.85 g; percent yield = 100%
B) Theoretical yield = 9.95 g; percent yield = 88.9%
C) Theoretical yield = 12.1 g; percent yield = 73.1%
D) Theoretical yield = 9.95 g; percent yield = 112%
▶️ Answer/Explanation
Ans: B
Step 1: Convert grams of Mg to moles
$n_{\text{Mg}} = \frac{6.00\,\text{g}}{24.3\,\text{g/mol}} = 0.2469\,\text{mol}$
Step 2: Use stoichiometry to find moles of MgO
From the equation, the mole ratio Mg : MgO = 2 : 2 = 1 : 1.
$n_{\text{MgO}} = 0.2469\,\text{mol}$
Step 3: Calculate theoretical yield in grams
$\text{Theoretical yield} = 0.2469 \times 40.3 = 9.95\,\text{g}$
Step 4: Calculate percent yield
$\% \text{ yield} = \frac{8.85}{9.95} \times 100 = 88.9\%$
Therefore, the correct answer is B — Theoretical yield = 9.95 g; percent yield = 88.9%.
Aspirin (C9H8O4, Mr = 180) is synthesised from salicylic acid (Mr = 138) in a 1 : 1 molar ratio:
C7H6O3 + C4H6O3 → C9H8O4 + CH3COOH
A student starts with 13.8 g of salicylic acid (Mr = 138) as the limiting reactant. The reaction has a known percent yield of 72.0%.
What mass of aspirin is actually collected by the student?
A) 12.96 g
B) 18.0 g
C) 25.0 g
D) 9.34 g
▶️ Answer/Explanation
Ans: A
Step 1: Moles of salicylic acid (limiting reactant)
$n = \frac{13.8\,\text{g}}{138\,\text{g/mol}} = 0.100\,\text{mol}$
Step 2: Theoretical yield of aspirin
Mole ratio salicylic acid : aspirin = 1 : 1, so moles of aspirin = 0.100 mol.
$\text{Theoretical yield} = 0.100\,\text{mol} \times 180\,\text{g/mol} = 18.0\,\text{g}$
Step 3: Apply percent yield to find actual yield
$\text{Actual yield} = \text{Theoretical yield} \times \frac{\%\text{ yield}}{100} = 18.0 \times \frac{72.0}{100} = 12.96\,\text{g}$
Therefore, the correct answer is A — 12.96 g.
Sodium hydroxide reacts with phosphoric acid to form sodium phosphate and water:
3NaOH + H3PO4 → Na3PO4 + 3H2O
A chemist needs to produce exactly 41.0 g of Na3PO4 (Mr = 164). The reaction has a percent yield of 82.0%. NaOH has Mr = 40.0.
What minimum mass of NaOH must the chemist start with to obtain the required 41.0 g of Na3PO4?
A) 30.0 g
B) 36.6 g
C) 44.7 g
D) 73.2 g
▶️ Answer/Explanation
Ans: D
This question works backwards from the desired actual yield through percent yield to find the required starting mass.
Step 1: Find the required theoretical yield of Na3PO4
Since percent yield = 82.0% and actual yield needed = 41.0 g:
$\text{Theoretical yield} = \frac{\text{Actual yield}}{\%\text{ yield}} \times 100 = \frac{41.0}{82.0} \times 100 = 50.0\,\text{g}$
Step 2: Convert theoretical yield to moles of Na3PO4
$n_{\text{Na}_3\text{PO}_4} = \frac{50.0\,\text{g}}{164\,\text{g/mol}} = 0.3049\,\text{mol}$
Step 3: Use stoichiometry to find moles of NaOH needed
Mole ratio NaOH : Na3PO4 = 3 : 1
$n_{\text{NaOH}} = 0.3049 \times 3 = 0.9146\,\text{mol}$
Step 4: Convert moles of NaOH to grams
$m_{\text{NaOH}} = 0.9146 \times 40.0 = 36.6\,\text{g}$
Therefore, the correct answer is B — 36.6 g NaOH.
Copper reacts with silver nitrate solution in a single displacement reaction:
Cu + 2AgNO3 → Cu(NO3)2 + 2Ag
12.7 g of Cu (Mr = 63.5) is added to a solution containing 27.2 g of AgNO3 (Mr = 170). The reaction proceeds with a percent yield of 85.0%.
What is the actual mass of silver (Mr = 108) deposited?
A) 27.5 g
B) 23.4 g
C) 14.9 g
D) 17.5 g
▶️ Answer/Explanation
Ans: B
Step 1: Convert to moles
$n_{\text{Cu}} = \frac{12.7}{63.5} = 0.200\,\text{mol}$ $n_{\text{AgNO}_3} = \frac{27.2}{170} = 0.160\,\text{mol}$
Step 2: Identify the limiting reactant
Ratio: 1 mol Cu requires 2 mol AgNO3.
For 0.200 mol Cu: need \(0.200 \times 2 = 0.400\) mol AgNO3.
Available AgNO3 = 0.160 mol → far less than needed.
AgNO3 is the limiting reactant.
Step 3: Calculate theoretical yield of Ag
Ratio: 2 mol AgNO3 → 2 mol Ag (1 : 1 ratio)
$n_{\text{Ag}} = 0.160\,\text{mol}$ $\text{Theoretical yield} = 0.160 \times 108 = 17.28\,\text{g}$
Step 4: Apply percent yield
$\text{Actual yield} = 17.28 \times \frac{85.0}{100} = 14.69 \approx 14.7\,\text{g}$
Therefore, the correct answer is C — 14.9 g.