Concept:Combustion stoichiometry allows us to directly relate the mass of a fuel to the mass of oxygen required to burn it, provided we use the balanced equation and consistent units (kg can be used directly as kmoles).
Formula:$$ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} $$
Solution:- The balanced equation reveals that 1 mole of Methane (\( \text{CH}_4 \)) requires exactly 2 moles of Oxygen (\( \text{O}_2 \)).
- Molar mass of \( \text{CH}_4 = 16 \text{ g/mol} \). Molar mass of \( 2\text{O}_2 = 2 \times 32 = 64 \text{ g} \).
- So, 16g of \( \text{CH}_4 \) requires 64g of \( \text{O}_2 \). (This is a 1:4 mass ratio).
- Therefore, 3.2 kg of methane will require exactly 4 times its mass in oxygen.
- \( 3.2 \text{ kg} \times 4 = 12.8 \text{ kg of Oxygen} \).
Why other options are incorrect:- 3.2 kg: Assumes a 1:1 mass ratio, ignoring the chemistry completely.
- 6.4 kg: Assumes a 1:2 mass ratio (which is the molar ratio, not the mass ratio).
- 15.4 kg: Mathematically arbitrary.
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