U4.5 · StoichiometryPractice · 化学计量

n($\ce{KClO3}$) = 12.25 ÷ 122.55 = 0.09996 mol. The mole ratio of $\ce{KClO3}$ to $\ce{O2}$ is 2 : 3, so n($\ce{O2}$) = 0.09996 × 3/2 = 0.1499 mol and m = 0.1499 × 32.00 = 4.80 g. A 1 : 1 ratio would give 3.20 g.

n(NaOH) = 10.0 ÷ 40.00 = 0.2500 mol and n(HCl) = 10.0 ÷ 36.46 = 0.2743 mol. They react 1 : 1, so the smaller amount, NaOH, is used up and HCl is in excess. HCl left: 0.2743 − 0.2500 = 0.0243 mol, which is 0.0243 × 36.46 = 0.886 g. Equal masses don’t mean equal moles.

Theoretical yield: n($\ce{CaCO3}$) = 25.0 ÷ 100.09 = 0.2498 mol, giving 0.2498 × 56.08 = 14.01 g of CaO (1 : 1). Percent yield = actual ÷ theoretical = 12.6 ÷ 14.01 × 100% = 89.9%.

Mole ratios equal coefficient ratios: n($\ce{H2O}$) : n($\ce{O2}$) = 6 : 5, so n($\ce{H2O}$) = 0.60 × 6/5 = 0.72 mol. 0.90 mol uses the $\ce{H2O}$ : $\ce{NH3}$ ratio, 6 : 4, by mistake; 0.50 mol inverts the ratio.

n($\ce{C3H8}$) = 4.40 ÷ 44.09 = 0.09980 mol, n($\ce{O2}$) = 0.09980 × 5 = 0.4990 mol, and V = 0.4990 × 22.4 = 11.2 L.

Before the break, each mL adds the same mass: Pb²⁺ is limiting, so every bit added precipitates. The break is at 50 mL: n(Pb²⁺) = 0.100 × 0.0500 = 0.00500 mol, exactly half of n(I⁻) = 0.0100 mol (Pb²⁺ + 2I⁻ → PbI₂). After that the I⁻ is gone, and the mass stays at 0.00500 × 461.0 ≈ 2.31 g.

The plateau means the acid is used up: at most 560 mL of H₂, or 0.560 L ÷ 22.4 L/mol = 0.0250 mol. From Mg + 2HCl → MgCl₂ + H₂, n(HCl) = 2 × 0.0250 = 0.0500 mol, and [HCl] = 0.0500 ÷ 0.0500 L = 1.00 M. The break is at about 0.61 g of Mg (0.0250 mol); magnesium added after that is in excess. Choosing 0.500 M forgets the factor of 2.