Moles and Avogadro's Numbers

Moles of O₂ = 64 g ÷ 32 g/mol = 2.0 moles. At STP, molar volume is 22.4 dm³. Therefore, volume = 2.0 × 22.4 = 44.8 dm³.

Moles of CO₂ = 88 ÷ 44 = 2.0 moles. Molecules = 2.0 × 6.022 × 10²³ = 1.2044 × 10²⁴.

The mole ratio Na : NaCl is 1:1. Therefore, 0.5 moles of Na produce 0.5 moles of NaCl. Mass = 0.5 × 58.5 = 29.25 g.

Moles of N₂ = 28 ÷ 28 = 1 mole. One mole of N₂ forms 2 moles of NH₃. Volume = 2 × 22.4 = 44.8 dm³.

Moles = 1 ÷ 22.4 = 0.0446 mol. Molar mass = 1.25 ÷ 0.0446 ≈ 28 g/mol.

Each AlCl₃ gives 4 ions. Therefore, 0.5 × 4 = 2 moles of ions. Number of ions = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴.

The volume ratio SO₂ : SO₃ is 1:1. Hence, 44.8 dm³ of SO₂ produces 44.8 dm³ of SO₃.

Molar mass = (4.4 × 10⁻²² g) × (6.022 × 10²³ mol⁻¹) ≈ 265 g/mol.

30 g of C₂H₆ = 1 mole. One mole of C₂H₆ produces 3 moles of H₂O. Mass of water = 3 × 18 = 54 g.

44.8 dm³ = 2 moles NH₃. Each molecule has 4 atoms. Total atoms = 8 moles of atoms = 4.8176 × 10²⁴ atoms.

50 g CaCO₃ = 0.5 mole. It forms 0.5 mole CaO. Mass = 0.5 × 56 = 28 g.

108 g Al = 4 moles. The ratio 4Al : 2Al₂O₃ gives 2 moles Al₂O₃.

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

1.6 g O₂ = 0.05 mol. Molecules = 0.05 × 6.022 × 10²³ = 3.011 × 10²².

Molecular formula is C₆H₁₂O₆ because 180/30 = 6. Thus each mole contains 12 moles hydrogen atoms.

Gas volumes follow mole ratios. Ratio N₂:H₂ = 1:3. Therefore 5 dm³ nitrogen requires 15 dm³ hydrogen.

One mole pentane forms 5 CO₂ + 6 H₂O = 11 mol products. Therefore 0.25 mol forms 2.75 mol products.

Multiplying by Avogadro's number gives approximately 56 g/mol, corresponding to iron.

Iron = 0.5 mol, sulphur = 0.625 mol. Since the reaction ratio is 1:1, iron limits the reaction.