MCQs

One mole of Fe₂O₃ reacts with 3 mol CO to produce 2 mol Fe = 112 g.

H₂ is limiting. NH₃ formed = (2/3) × 4 = 2.67 mol.

Al = 2 mol, Cl₂ = 1 mol. Two moles Al require 3 mol Cl₂, so chlorine is limiting.

The reaction is 1:1. Two moles of O₂ produce only 2 moles of CO₂.

8 mol O₂ require only 6.4 mol NH₃. Therefore, NH₃ remains in excess by 1.6 mol.

The ratio 10:15 simplifies to 2:3, matching the balanced equation. Neither reactant is in excess.

4 g H₂ = 2 mol and 32 g O₂ = 1 mol. The ratio is exactly 2:1, so both reactants are completely consumed.

The limiting reactant is completely used up first and determines the maximum amount of product formed.

According to the balanced equation, 2 moles of N₂ require 6 moles of H₂. Since only 3 moles of H₂ are available, hydrogen is the limiting reactant.

9 g H₂O = 0.5 mole. Each molecule contains 10 protons. Total protons = 0.5 × 10 × Nₐ = 3.011 × 10²⁴.

49 g H₂SO₄ = 0.5 mole. NaOH:H₂SO₄ = 2:1. Therefore, NaOH required = 1 mole = 40 g.

One mole occupies 22.4 dm³. Therefore, molecules in 1 dm³ = (6.022 × 10²³) ÷ 22.4 ≈ 2.69 × 10²².

Number of moles = 6.022 × 10²¹ ÷ 6.022 × 10²³ = 0.01 mole. Atomic mass = 0.64 ÷ 0.01 = 64 g/mol (64 u).

The volume ratio H₂:Cl₂ is 1:1. Therefore, 11.2 dm³ of hydrogen requires 11.2 dm³ of chlorine.

Moles of O₂ = 9.6 ÷ 32 = 0.3 mole. Required KClO₃ = 0.3 × 2/3 = 0.2 mole. Mass = 0.2 × 122.5 = 24.5 g.

22 g CO₂ = 0.5 mole. One mole of CO₂ contains one mole of carbon atoms. Mass of carbon = 0.5 × 12 = 6 g.

1.7 g NH₃ = 0.1 mole. Each NH₃ molecule has 10 electrons. Therefore, total electrons = 0.1 × 10 × Nₐ = 6.022 × 10²³.

Mole ratio S:H₂S = 3:2. Moles of S = 0.5 × 3/2 = 0.75 mole. Mass = 0.75 × 32 = 24 g.

Moles of O₂ = 16 ÷ 32 = 0.5 mole. By Avogadro's law, equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Therefore, the same volume of CO₂ also contains 0.5 mole.