Chemistry · Ch 1 — Basic Concepts of Chemistry and Chemical Calculations
Stoichiometric Calculations
Stoichiometric Calculations
Stoichiometry is formally the quantitative mole relationship between reactants and products in a balanced chemical equation. In practice, though, a reactant or product amount might be given to you in moles, in mass, or in gas volume -- and the great convenience of stoichiometry is that these three units are all freely inter-convertible:
- moles → mass: multiply by molar mass
- mass → moles: divide by molar mass
- moles → volume (of a gas, at 0°C and 1 atm): multiply by 22.4 L
- volume → moles: divide by 22.4 L
Table 1.7.1 works through all of these conversions together for one reaction, the combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g): the mole-mole relationship (1 : 2 : 1 : 2) converts to a mass-mass relationship (16 g : 64 g : 44 g : 36 g, using each substance's molar mass) and to a volume-volume relationship (22.4 L : 44.8 L : 22.4 L : 44.8 L, using 22.4 L per mole of gas at 273 K/1 atm).
Five further worked calculations consolidate the method:
- Example 1.6: From N₂ + 3H₂ → 2NH₃, producing 10 moles of ammonia needs (3/2) × 10 = 15 moles of hydrogen.
- Example 1.7: From CH₄ + 2O₂ → CO₂ + 2H₂O, burning 32 g of methane (i.e. 2 moles) produces (36/16) × 32 = 72 g of water.
- Example 1.8: From CaCO₃ →(Δ) CaO + CO₂, heating 50 g of calcium carbonate completely gives (22.7/100) × 50 = 11.35 L of CO₂ at STP.
- Example 1.9: From H₂ + Cl₂ → 2HCl, forming 11.2 L of HCl at 273 K/1 atm needs (22.4/44.8) × 11.2 = 5.6 L of chlorine. …
| CH₄(g) | O₂(g) | CO₂(g) | H₂O(g) | |
|---|---|---|---|---|
| Stoichiometric coefficient | 1 | 2 | 1 | 2 |
| Mole-mole | 1 mole | 2 moles | 1 mole | 2 moles |
| Mass-mass (mol × molar mass) | 1×16 = 16 g | 2×32 = 64 g | 1×44 = 44 g | 2×18 = 36 g |
| Mass-volume (volume of product at 273 K, 1 atm) | 16 g | 64 g | 22.4 L | 44.8 L |
Worked out. For N₂(g) + 3H₂(g) → 2NH₃(g), since 2 moles of ammonia need 3 moles of hydrogen, producing 10 moles of ammonia needs (3/2) × 10 = 15 moles of hydrogen. …
Worked out. For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), 1 mole (16 g) of CH₄ produces 2 moles (36 g) of water, so combustion of 32 g of CH₄ produces (36/16) × 32 = 72 g of water. …
Worked out. For CaCO₃(s) →(Δ) CaO(s) + CO₂(g), 1 mole (100 g) of CaCO₃ produces 1 mole of CO₂, which occupies 22.7 L at STP; so 50 g of CaCO₃ produces (22.7/100) × 50 = 11.35 L of CO₂. …
Worked out. For H₂(g) + Cl₂(g) → 2HCl(g) at 273 K and 1 atm, 1 mole (22.4 L) of Cl₂ produces 2 moles (44.8 L) of HCl; so producing 11.2 L of HCl needs (22.4/44.8) × 11.2 = 5.6 L of chlorine. …
Worked out. For MgCO₃ →(Δ) MgO + CO₂ (molar mass of MgCO₃ = 84 g mol⁻¹, containing 24 g Mg, 12 g C and 48 g O per 84 g), the percentage composition works out to Mg = 28.57%, C = 14.29% and O = 57.14%. Since 84 g of 100%-pure MgCO₃ gives 44 g of CO₂ on heating, 1000 g of 90%-pure MgCO₃ gives (44/84) × 0.90 × 1000 = 471.43 g = 0.471 kg of CO₂. …