2.6 · The Mole and StoichiometryPractice · 摩尔与化学计量

Each $\ce{C2H6}$ molecule has 2 C and 6 H, 8 atoms in all, so the moles of atoms = 0.200 mol × 8 = 1.60 mol, and the number of atoms = 1.60 × $6.022 \times 10^{23}$ = $9.64 \times 10^{23}$. The question asks for atoms, not molecules (there are only 0.200 mol of molecules).

First the moles: n = N ÷ $N_A$ = $3.01 \times 10^{22}$ ÷ $6.022 \times 10^{23}$ = 0.04998 mol. Then the mass: m = n × M = 0.04998 mol × 64.06 g/mol = 3.20 g.

Convert everything to moles of O atoms: 0.30 mol $\ce{H2SO4}$: 1.20 mol O; 1.0 mol $\ce{H2O}$: 1.00 mol O; 44 g $\ce{CO2}$: 2.00 mol O; 16 g $\ce{O2}$: 1.00 mol O. 44 g of $\ce{CO2}$ is about 1 mol, and each molecule has 2 O, so it has the most. Put everything in moles first, then multiply by the O atoms per formula.

The molar mass of $\ce{CaCO3}$ is 40.08 + 12.01 + 3 × 16.00 = 100.09 g/mol, of which Ca is 40.08 g. Mass percent = 40.08 ÷ 100.09 × 100% = 40.0%.

Take 100 g: n(C) = 82.7 ÷ 12.01 = 6.886 mol and n(H) = 17.3 ÷ 1.008 = 17.16 mol. Divide by the smaller: C : H = 1 : 2.49, not close to a whole number, so multiply both by 2 to get 2 : 5: empirical formula $\ce{C2H5}$ (29.06 g/mol). 58.12 ÷ 29.06 ≈ 2, so the molecular formula is $\ce{C4H10}$ (butane). $\ce{C2H5}$ is only the empirical formula.

n($\ce{C2H5OH}$) = 6.90 g ÷ 46.07 g/mol = 0.1498 mol. The equation gives a 1 : 3 mole ratio, so n($\ce{O2}$) = 3 × 0.1498 = 0.4494 mol and m($\ce{O2}$) = 0.4494 mol × 32.00 g/mol = 14.4 g. Coefficients give mole ratios, not mass ratios.

n(Fe) = 5.60 ÷ 55.85 = 0.100 mol and n(S) = 4.00 ÷ 32.06 = 0.125 mol. They react 1 : 1 and there is less Fe, so Fe is limiting and S is in excess. Moles of FeS = moles of Fe, so m(FeS) = 5.60 g × 87.91 ÷ 55.85 = 8.81 g. Simply adding the two masses (9.60 g) is wrong, because not all the sulfur reacts.

The sulfur that reacts matches the iron mole for mole: m(S) = 5.60 g × 32.06 ÷ 55.85 = 3.21 g. Sulfur left = 4.00 − 3.21 = 0.79 g. Check: 8.81 g of FeS + 0.79 g of leftover S = the original 9.60 g, so mass is conserved.

Theoretical yield: n($\ce{CaCO3}$) = 10.0 ÷ 100.09 = 0.09991 mol, and the same moles of CaO form, so the theoretical yield is 0.09991 × 56.08 = 5.603 g. Percent yield = actual ÷ theoretical × 100% = 5.04 ÷ 5.603 × 100% = 90.0%. Dividing by the reactant’s mass (5.04 ÷ 10.0) is wrong.

Coefficients give mole ratios: $\ce{O2}$ : $\ce{Al2O3}$ = 3 : 2, so n($\ce{Al2O3}$) = 0.60 mol × 2/3 = 0.40 mol. 0.90 mol comes from flipping the ratio (× 3/2).