How much? The amount of chemical change practice questions.

For the reaction 2H₂ + O₂ → 2H₂O, 4 mol of H₂ is mixed with 1 mol of O₂. Which is the limiting reagent?

• hydrogen
• oxygen
• water
• neither – they are in exact ratio

Answer: oxygen

The ratio needed is 2 H₂ : 1 O₂. 1 mol O₂ would react with only 2 mol H₂, so O₂ runs out first while 2 mol H₂ remains in excess.

Determine the number of moles in 5.85 g of sodium chloride, NaCl. (M = 58.44 g mol⁻¹.)

• 0.100 mol
• 0.585 mol
• 10.0 mol
• 0.0585 mol

Answer: 0.100 mol

n = mass/M = 5.85/58.44 = 0.100 mol.

In the reaction Fe₂O₃ + 3CO → 2Fe + 3CO₂, determine the amount of iron produced from 0.50 mol of Fe₂O₃ (assume CO is in excess).

• 1.0 mol
• 0.50 mol
• 2.0 mol
• 0.25 mol

Answer: 1.0 mol

The mole ratio Fe₂O₃ : Fe is 1 : 2, so 0.50 mol Fe₂O₃ produces 0.50 × 2 = 1.0 mol Fe.

25.0 cm³ of 0.200 mol dm⁻³ NaOH is exactly neutralised by HCl. Calculate the amount, in mol, of HCl required (NaOH + HCl → NaCl + H₂O).

• 5.00 × 10⁻³ mol
• 0.200 mol
• 5.00 mol
• 2.50 × 10⁻³ mol

Answer: 5.00 × 10⁻³ mol

n(NaOH) = c × V = 0.200 × 0.0250 = 5.00 × 10⁻³ mol; 1 : 1 ratio means n(HCl) = 5.00 × 10⁻³ mol.

Identify the correct statement about the percentage atom economy of a reaction.

• It is fixed by the balanced equation, not by the practical technique
• It is measured by comparing the actual yield with the theoretical yield
• It falls whenever product is lost during filtration or transfer
• It is always numerically equal to the reaction's percentage yield

Answer: It is fixed by the balanced equation, not by the practical technique

Atom economy compares the mass of the desired product with the total mass of the reactants and can be worked out from the equation alone, before any experiment is done. Percentage yield is the experimental quantity, so losses in the laboratory change the yield but never the atom economy.

0.20 mol of N₂(g) and 0.20 mol of H₂(g) are mixed and react according to N₂(g) + 3H₂(g) → 2NH₃(g). Determine the limiting reactant and the maximum amount of ammonia formed.

• H₂, giving 0.133 mol of NH₃
• N₂, giving 0.400 mol of NH₃
• H₂, giving 0.200 mol of NH₃
• N₂, giving 0.133 mol of NH₃

Answer: H₂, giving 0.133 mol of NH₃

0.20 mol of N₂ would need 0.60 mol of H₂, but only 0.20 mol is available, so hydrogen is limiting and n(NH₃) = 0.20 × 2/3 = 0.133 mol. Equal starting amounts do not mean the reactants run out together — the coefficients decide.