Home / AP® Exam / Pre AP Chemistry / Pre AP Chemistry

Pre AP Chemistry -3.2C Limiting Reactants- MCQ Exam Style Questions -New Syllabus 2025-2026

Pre AP Chemistry -3.2C Limiting Reactants- MCQ Exam Style Questions – New Syllabus 2025-2026

Pre AP Chemistry -3.2C Limiting Reactants- 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 reacts with dilute hydrochloric acid.

The equation for the reaction is shown.

CaCO3 + 2HCl → CaCl2 + H2O + CO2

1.00 g of calcium carbonate is added to 50.0 cm3 of 0.0500 mol / dm3 hydrochloric acid.

Which volume of carbon dioxide is made in this reaction?

A) 30 cm3        
B) 60 cm3        
C) 120 cm3        
D) 240 cm3

▶️ Answer/Explanation
Solution

Ans: A

Step 1: Calculate moles of reactants
– Moles of CaCO3 = \(\frac{1.00\,g}{100.1\,g/mol} = 0.0100\,mol\)
– Moles of HCl = \(0.0500\,mol/dm^3 \times 0.0500\,dm^3 = 0.00250\,mol\)

Step 2: Determine limiting reactant
– The reaction requires 2 moles HCl per 1 mole CaCO3.
– Available HCl can react with \(\frac{0.00250}{2} = 0.00125\,mol\) CaCO3.
– HCl is the limiting reactant.

Step 3: Calculate moles of CO2 produced
– From the equation, 2 moles HCl produce 1 mole CO2.
– Moles of CO2 = \(\frac{0.00250}{2} = 0.00125\,mol\)

Step 4: Convert to volume at room conditions
– Molar volume of gas ≈ 24,000 cm3/mol at room temperature.
– Volume of CO2 = \(0.00125\,mol \times 24,000\,cm^3/mol = 30\,cm^3\).

Therefore, the correct answer is A (30 cm3).

Question 

Calcium carbonate reacts with dilute hydrochloric acid according to the equation shown.

\( CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O \)

10 g of calcium carbonate is reacted with 100 cm3 of 1 mol/dm3 hydrochloric acid.

The following statements are made.

  1. 1.2 dm3 of carbon dioxide is formed.
  2. 5.6 g of calcium chloride is formed.
  3. 4.8 g of carbon dioxide is formed.
  4. No calcium carbonate is left when the reaction is completed.

Which statements about the reaction are correct?

A) 1 and 2
B) 1 and 4
C) 2 and 3
D) 3 and 4

▶️ Answer/Explanation
Solution

Ans: A (1 and 2)

Step 1: Determine limiting reactant

  • Moles of \( CaCO_3 = \frac{10\,g}{100\,g/mol} = 0.1\,mol \)
  • Moles of \( HCl = 1\,mol/dm^3 \times 0.1\,dm^3 = 0.1\,mol \)

Since the reaction requires 2 moles of HCl per mole of \( CaCO_3 \), HCl is the limiting reactant (only 0.05 mol \( CaCO_3 \) can react).

Step 2: Calculate products formed

  • CO2 volume: \[ 0.05\,mol \times 24\,dm^3/mol = 1.2\,dm^3 \] (Statement 1 correct)
  • CaCl2 mass: \[ 0.05\,mol \times 111\,g/mol = 5.55\,g \approx 5.6\,g \] (Statement 2 correct)
  • CO2 mass: \[ 0.05\,mol \times 44\,g/mol = 2.2\,g \] (Statement 3 incorrect)

Step 3: Check excess reactant

Only 0.05 mol of \( CaCO_3 \) reacts, leaving \( 0.1 – 0.05 = 0.05\,mol \) unreacted (Statement 4 incorrect).

Question 

A student mixed together 25.0 cm3 of 1.00 mol/dm3 hydrochloric acid and 25.0 g of calcium carbonate.

2HCl(aq) + CaCO3(s) → CaCl2(aq) + H2O(l) + CO2(g)

What is the maximum volume of carbon dioxide gas that could be collected at room temperature and pressure?

A) 300 dm3
B) 6.00 dm3
C) 0.600 dm3
D) 0.300 dm3

▶️ Answer/Explanation
Solution

Ans: D

1. Moles of HCl: \[ n = c \times V = 1.00\, \text{mol/dm}^3 \times 0.0250\, \text{dm}^3 = 0.0250\, \text{mol} \] 2. Moles of CaCO3: \[ n = \frac{m}{M} = \frac{25.0\, \text{g}}{100.1\, \text{g/mol}} = 0.250\, \text{mol} \] 3. Limiting reactant: The reaction requires 2 moles of HCl per 1 mole of CaCO3. – HCl is limiting (0.0250 mol HCl can react with 0.0125 mol CaCO3). 4. Moles of CO2: 0.0250 mol HCl produces 0.0125 mol CO2 (1:1 ratio with CaCO3). 5. Volume of CO2 at RTP: \[ V = n \times 24.0\, \text{dm}^3/\text{mol} = 0.0125\, \text{mol} \times 24.0\, \text{dm}^3/\text{mol} = 0.300\, \text{dm}^3 \] Thus, the maximum volume of CO2 is 0.300 dm3 (Option D).

Question 

Which quantities of chemicals will react exactly with no reactants left over?

A) 12 g of carbon and 12 g of oxygen
B) 12 g of carbon and 48 g of oxygen
C) 12 g of magnesium and 16 g of oxygen
D) 24 g of magnesium and 16 g of oxygen

▶️ Answer/Explanation
Solution

Ans: D

To determine which combination reacts completely:

  1. For Option D (Magnesium + Oxygen):
    • Balanced equation: \( 2Mg + O_2 \rightarrow 2MgO \)
    • Molar masses: \( Mg = 24 \, \text{g/mol}, O_2 = 32 \, \text{g/mol} \)
    • Moles of Mg: \( \frac{24 \, \text{g}}{24 \, \text{g/mol}} = 1 \, \text{mol} \)
    • Moles of \( O_2 \): \( \frac{16 \, \text{g}}{32 \, \text{g/mol}} = 0.5 \, \text{mol} \)
    • Stoichiometric ratio: 2:1 (Mg:\( O_2 \)). Here, 1 mol Mg reacts exactly with 0.5 mol \( O_2 \).
  2. Why other options are incorrect:
    • A: \( C + O_2 \rightarrow CO_2 \). 12 g C (1 mol) requires 32 g \( O_2 \), not 12 g.
    • B: 12 g C (1 mol) reacts with 32 g \( O_2 \), leaving excess \( O_2 \).
    • C: 12 g Mg (0.5 mol) requires 0.25 mol \( O_2 \) (8 g), not 16 g.

Only Option D has stoichiometrically exact quantities (24 g Mg + 16 g \( O_2 \)).

Scroll to Top