MCQs

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.

0.5 × 6 = 3 moles of oxygen atoms. Therefore, the oxygen subscript is 6.

2 moles C₂H₆ require 7 moles O₂. Mass = 7 × 32 = 224 g.

32.5 g Zn = 0.5 mole. It gives 0.5 mole H₂. Volume = 0.5 × 22.4 = 11.2 dm³.

143.5 g = 0.5 mole washing soda. Water = 0.5 × 10 = 5 moles.

Fe:H₂ = 3:4. For 3 moles H₂, Fe = 2.25 moles. Mass = 2.25 × 56 = 126 g.

Molar mass = Density × 22.4 = 1.964 × 22.4 ≈ 44 g/mol.

3.011 × 10²³ molecules = 0.5 mole. Volume = 0.5 × 22.4 = 11.2 dm³.