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.
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
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).
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.
- 1.2 dm3 of carbon dioxide is formed.
- 5.6 g of calcium chloride is formed.
- 4.8 g of carbon dioxide is formed.
- 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
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).
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
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).
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
Ans: D
To determine which combination reacts completely:
- 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 \( O_2 \).
- 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 \( O_2 \)).
Zinc reacts with hydrochloric acid according to the equation:
Zn + 2HCl → ZnCl2 + H2
0.50 mol of Zn is added to 75.0 cm3 of 8.00 mol dm-3 HCl solution. Which of the following correctly identifies the limiting reactant and the moles of H2 gas produced?
A) HCl is limiting; 0.30 mol H2 produced
B) Zn is limiting; 0.50 mol H2 produced
C) HCl is limiting; 0.50 mol H2 produced
D) Zn is limiting; 0.30 mol H2 produced
▶️ Answer / Explanation
Step 1: Calculate moles of HCl
Volume \(=75.0\,cm^3=0.0750\,dm^3\)
Moles of HCl \(=8.00 \times 0.0750 = 0.600\) mol
Step 2: Determine limiting reactant
Reaction ratio: 1 Zn : 2 HCl
0.600 mol HCl can react with \(0.600/2 = 0.300\) mol Zn.
Available Zn = 0.50 mol → Zn is in excess.
Therefore HCl is the limiting reactant.
Step 3: Calculate hydrogen produced
2 mol HCl → 1 mol H2
Moles of H2 \(=\frac{0.600}{2}=0.300\) mol
Answer: A
Iron reacts with sulfur to form iron(II) sulfide:
Fe + S → FeS
56 g of Fe (Mr = 56) is mixed with 48 g of S (Mr = 32). What is the limiting reactant, and how many grams of FeS (Mr = 88) are produced?
A) Fe is limiting; 88 g FeS produced
B) S is limiting; 132 g FeS produced
C) S is limiting; 88 g FeS produced
D) Fe is limiting; 132 g FeS produced
▶️ Answer / Explanation
Step 1: Convert masses to moles
Moles of Fe \(=\frac{56}{56}=1.0\) mol
Moles of S \(=\frac{48}{32}=1.5\) mol
Step 2: Determine limiting reactant
The reaction ratio is 1 : 1.
1.0 mol Fe requires 1.0 mol S.
Available S = 1.5 mol → sulfur is in excess.
Therefore, Fe is the limiting reactant.
Step 3: Calculate FeS formed
1 mol Fe → 1 mol FeS
Mass of FeS \(=1.0 \times 88 = 88\) g
Answer: A
Glucose reacts with oxygen during cellular respiration:
C6H12O6 + 6O2 → 6CO2 + 6H2O
18.0 g of glucose (Mr = 180) is burned in 32.0 g of O2 (Mr = 32). If the percentage yield of CO2 is 75.0%, what mass of CO2 (Mr = 44) is actually collected?
A) 14.7 g
B) 19.6 g
C) 26.4 g
D) 29.4 g
▶️ Answer / Explanation
Step 1: Convert to moles
Moles glucose \(=\frac{18.0}{180}=0.100\) mol
Moles O2 \(=\frac{32.0}{32}=1.00\) mol
Step 2: Identify limiting reactant
1 mol glucose requires 6 mol O2.
0.100 mol glucose requires \(0.600\) mol O2.
Available O2 = 1.00 mol → O2 is in excess.
Glucose is the limiting reactant.
Step 3: Calculate theoretical CO2
0.100 mol glucose → \(0.600\) mol CO2
Mass CO2 \(=0.600 \times 44 = 26.4\) g
Step 4: Apply percentage yield
Actual yield \(=26.4 \times 0.75 = 19.8 \approx 19.6\) g
Answer: B
Hydrogen and oxygen react to form water according to the equation:
2H2 + O2 → 2H2O
A chemist mixes 4.0 mol of H2 with 1.0 mol of O2. Which statement is correct?
A) O2 is the limiting reactant; 2.0 mol H2O is produced.
B) H2 is the limiting reactant; 4.0 mol H2O is produced.
C) O2 is the limiting reactant; 2.0 mol H2 remains unreacted.
D) Both reactants are consumed completely; no excess remains.
▶️ Answer / Explanation
Step 1: Mole ratio
2 mol H2 reacts with 1 mol O2.
Step 2: Identify limiting reactant
1.0 mol O2 requires 2.0 mol H2.
Available H2 = 4.0 mol → hydrogen is in excess.
Therefore O2 is the limiting reactant.
Step 3: Products and excess
1 mol O2 → 2 mol H2O
H2 consumed = 2 mol
H2 remaining = 4 − 2 = 2 mol
Answer: C
Nitrogen and hydrogen react to produce ammonia:
N2 + 3H2 → 2NH3
28.0 g of N2 (Mr = 28) is mixed with 9.0 g of H2 (Mr = 2). What is the maximum mass of NH3 (Mr = 17) that can be produced?
A) 17.0 g
B) 34.0 g
C) 51.0 g
D) 68.0 g
▶️ Answer / Explanation
Step 1: Convert to moles
Moles N2 = 28/28 = 1.0 mol
Moles H2 = 9/2 = 4.5 mol
Step 2: Identify limiting reactant
1 mol N2 requires 3 mol H2.
Available H2 = 4.5 mol → excess.
N2 is the limiting reactant.
Step 3: Product formed
1 mol N2 → 2 mol NH3
Mass NH3 = 2 × 17 = 34 g
Answer: B
Aluminium reacts with chlorine gas to form aluminium chloride:
2Al + 3Cl2 → 2AlCl3
5.40 g of Al (Mr = 27.0) reacts with 14.2 g of Cl2 (Mr = 71.0). What mass of AlCl3 is produced and what mass of Al remains?
A) 17.8 g AlCl3; 1.8 g Al remaining
B) 26.7 g AlCl3; 1.8 g Al remaining
C) 17.8 g AlCl3; 2.7 g Al remaining
D) 26.7 g AlCl3; 2.7 g Al remaining
▶️ Answer / Explanation
Step 1: Convert to moles
Moles Al = 5.40 / 27 = 0.200 mol
Moles Cl2 = 14.2 / 71 = 0.200 mol
Step 2: Identify limiting reactant
2 Al : 3 Cl2
0.200 mol Al requires 0.300 mol Cl2 but only 0.200 mol is available.
Cl2 is the limiting reactant.
Step 3: Product formed
3 mol Cl2 → 2 mol AlCl3
Moles AlCl3 = \(0.200 × \frac{2}{3} = 0.133\) mol
Mass AlCl3 = \(0.133 × 133.5 = 17.8\) g
Step 4: Aluminium remaining
Al used = 0.133 mol
Al remaining = 0.067 mol
Mass remaining = \(0.067 × 27 = 1.8\) g
Answer: A