“Stoichiometry” MCQ with Answer Explanation.

Here some basic mcq’s about “stoichiometry” with answer which is explained in details. Let’s check one by one which is given below.

1. What is the molar mass of water (H₂O)?

A) 18 g/mol
B) 16 g/mol
C) 20 g/mol
D) 22 g/mol

Answer: A) 18 g/mol

Explanation: The molar mass of water is calculated as follows: H (1 g/mol) × 2 + O (16 g/mol) = 2 + 16 = 18 g/mol.

2. In the reaction 2H₂ + O₂ → 2H₂O, how many moles of H₂ are needed to completely react with 1 mole of O₂?

A) 1 mole
B) 2 moles
C) 3 moles
D) 4 moles

Answer: B) 2 moles

Explanation: According to the balanced equation, 2 moles of H₂ are required for every 1 mole of O₂.

3. What is the mass of 0.5 moles of sodium chloride (NaCl)? (Molar mass of NaCl = 58.5 g/mol)

A) 29.25 g
B) 58.5 g
C) 100 g
D) 117 g

Answer: A) 29.25 g

Explanation: Mass = moles × molar mass = 0.5 moles × 58.5 g/mol = 29.25 g.

4. If 3 moles of A react with 2 moles of B to form 4 moles of C, what is the stoichiometric ratio of A to C?

A) 3:4
B) 4:3
C) 2:3
D) 3:2

Answer: A) 3:4

Explanation: The reaction shows that 3 moles of A produce 4 moles of C, leading to a ratio of 3:4.

5. Which of the following statements is true about a limiting reactant?

A) It is always the reactant with the lowest molar mass.
B) It determines the amount of product formed.
C) It is completely consumed in the reaction.
D) Both B and C are correct.

Answer: D) Both B and C are correct.

Explanation: The limiting reactant is the one that runs out first, thus limiting the amount of product formed and is completely consumed during the reaction.

6. How many grams of CO₂ are produced when 10 g of glucose (C₆H₁₂O₆) are completely burned? (Molar mass of CO₂ = 44 g/mol)

A) 44 g
B) 88 g
C) 132 g
D) 176 g

Answer: B) 88 g

Explanation: The balanced equation for glucose combustion is C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O. From 1 mole of glucose, 6 moles of CO₂ are produced. The molar mass of glucose is 180 g/mol. Thus, 10 g of glucose produces (10 g / 180 g/mol) × 6 moles of CO₂ = 0.333 moles of CO₂, which is 0.333 moles × 44 g/mol = 14.67 g.

7. In a chemical reaction, if 4 moles of reactant A yield 8 moles of product B, what is the yield of B if 2 moles of A are used?

A) 2 moles
B) 4 moles
C) 6 moles
D) 8 moles

Answer: B) 4 moles

Explanation: The stoichiometry shows that 4 moles of A yield 8 moles of B. Thus, using 2 moles of A would yield (2 moles A) × (8 moles B / 4 moles A) = 4 moles of B.

8. What volume of 0.5 M HCl is required to neutralize 0.1 moles of NaOH? (Assume complete neutralization)

A) 100 mL
B) 200 mL
C) 400 mL
D) 500 mL

Answer: B) 200 mL

Explanation: The reaction between HCl and NaOH is 1:1. Therefore, to neutralize 0.1 moles of NaOH, 0.1 moles of HCl are required. Volume = moles / concentration = 0.1 moles / 0.5 M = 0.2 L or 200 mL.

9. What is the percentage yield if 10 g of product is obtained from 20 g of theoretical yield?

A) 50%
B) 75%
C) 100%
D) 125%

Answer: A) 50%

Explanation: Percentage yield = (actual yield / theoretical yield) × 100 = (10 g / 20 g) × 100 = 50%.

10. In the reaction N₂ + 3H₂ → 2NH₃, how many grams of NH₃ can be produced from 5 moles of H₂? (Molar mass of NH₃ = 17 g/mol)

A) 85 g
B) 34 g
C) 68 g
D) 51 g

Answer: C) 34 g

Explanation: 3 moles of H₂ produce 2 moles of NH₃. Therefore, 5 moles of H₂ can produce (5 moles H₂) × (2 moles NH₃ / 3 moles H₂) = 3.33 moles of NH₃. The mass of NH₃ produced = 3.33 moles × 17 g/mol = 56.61 g.