The Mole & Stoichiometry — Science

Moles = mass ÷ molar mass, so first build the molar mass of CaCO₃ by adding up its atoms.

Molar mass CaCO₃ = 40.08 (Ca) + 12.01 (C) + 3 × 16.00 (three O) g/mol.

That total is 100.09 g/mol; now divide 25.0 g by it.

0.25 mol

Molar mass of CaCO₃ = 40.08 + 12.01 + 3(16.00) = 100.09 g/mol.

Moles = mass ÷ molar mass = 25.0 g ÷ 100.09 g/mol.

= 0.250 mol of CaCO₃.

Follow the mass → moles → moles → mass road map: convert aluminium to moles before using any ratio.

Moles of Al = 53.96 g ÷ 26.98 g/mol. The mole ratio Fe : Al from the equation is 2 : 2, i.e. 1 : 1.

So moles of Fe = moles of Al; then multiply by 55.85 g/mol to get grams of iron.

111.7 g

Moles of Al = 53.96 ÷ 26.98 = 2.00 mol. The ratio Fe : Al is 2 : 2 = 1 : 1, so 2.00 mol of Fe forms.

Mass of Fe = 2.00 mol × 55.85 g/mol.

= 111.7 g of iron.

Use the coefficients to get moles of H₂ first, then convert those moles to grams.

The ratio H₂ : N₂ is 3 : 1, so 2.0 mol N₂ needs 2.0 × 3 mol of H₂.

Moles of H₂ = 6.0; grams = 6.0 × 2.02 g/mol.

12.12 g

Mole ratio H₂ : N₂ = 3 : 1, so 2.0 mol N₂ requires 2.0 × 3 = 6.0 mol H₂.

Convert to mass: 6.0 mol × 2.02 g/mol.

= 12.12 g of H₂.

First find the limiting reactant: compare the H₂ you have to the H₂ the nitrogen demands at a 3 : 1 ratio.

4.0 mol N₂ would need 12.0 mol H₂, but only 9.0 mol H₂ is present — so H₂ runs out first and controls the yield.

From the limiting H₂: moles NH₃ = 9.0 × (2 ÷ 3); then multiply by 17.04 g/mol.

102.24 g

Check the limiting reactant: 4.0 mol N₂ needs 4.0 × 3 = 12.0 mol H₂, but only 9.0 mol is available, so H₂ is limiting.

Moles of NH₃ = 9.0 mol H₂ × (2 NH₃ ÷ 3 H₂) = 6.0 mol; mass = 6.0 × 17.04 g/mol.

= 102.24 g of NH₃.

Percent yield = (actual ÷ theoretical) × 100, so first find the theoretical mass of CaO the 100.0 g of CaCO₃ should give.

Moles CaCO₃ = 100.0 ÷ 100.09 ≈ 1.00 mol; the ratio CaO : CaCO₃ is 1 : 1, so about 1.00 mol of CaO is expected.

Theoretical CaO ≈ 1.00 mol × 56.08 g/mol ≈ 56.0 g; now divide 44.8 by that and multiply by 100.

80 %

Moles of CaCO₃ = 100.0 ÷ 100.09 = 0.999 mol; at a 1 : 1 ratio that predicts 0.999 mol of CaO.

Theoretical mass = 0.999 × 56.08 ≈ 56.0 g of CaO. Percent yield = (44.8 ÷ 56.0) × 100.

≈ 80.0%.

The ratio H₂ : Cl₂ is 1 : 1, so whichever you have less of runs out first — that's the limiting reactant.

You have 2.0 mol Cl₂ and 3.0 mol H₂; Cl₂ is smaller, so Cl₂ is limiting and H₂ is in excess.

The 2.0 mol Cl₂ uses up 2.0 mol H₂ (1 : 1). Leftover H₂ = 3.0 − 2.0.

1 mol

At a 1 : 1 ratio, the 2.0 mol of Cl₂ is limiting and reacts with exactly 2.0 mol of H₂.

Hydrogen started at 3.0 mol, so leftover H₂ = 3.0 − 2.0.

= 1.0 mol of H₂ remaining.

Percent by mass of an element = (total mass of that element ÷ molar mass of the whole compound) × 100.

Molar mass of glucose = 6(12.01) + 12(1.01) + 6(16.00); the oxygen part is 6 × 16.00 = 96.00 g.

Divide 96.00 by the full molar mass (180.18), then × 100.

53.28 %

Molar mass of C₆H₁₂O₆ = 72.06 + 12.12 + 96.00 = 180.18 g/mol; oxygen contributes 6 × 16.00 = 96.00 g.

Percent oxygen = (96.00 ÷ 180.18) × 100.

= 53.28%.