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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.

Pre AP Chemistry – MCQ Exam Style Questions- All Topics

Question 

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
Solution

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%

Question 

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
Solution

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%).

Question 

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
Solution

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 (\( 2.3 \times 10^{-3}\, \text{mol/dm}^3 \)).

Question 

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
Solution

Ans: A

To determine the volume of ammonia produced:

  1. 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: \[ \frac{2}{3} \times 0.75 = 0.5 \text{ moles of } \mathrm{NH}_3. \]
  2. Account for 30% yield:
    • Actual yield = \(0.5 \times 0.30 = 0.15\) moles of \(\mathrm{NH}_3\).
  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 (\(3600 \mathrm{~cm}^3\)).

Question 

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
Solution

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%).

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