Concept:Stoichiometry allows us to calculate the mass of product generated from a known mass of reactant using the balanced chemical equation.
Formula:$$ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} $$
Solution:- First, find the moles of the limiting reactant (\( \text{CH}_4 \)):
- Molar mass of \( \text{CH}_4 = 12 + 4 = 16 \text{ g/mol} \).
- Moles of \( \text{CH}_4 = \frac{8 \text{ g}}{16 \text{ g/mol}} = 0.5 \text{ moles} \).
- From the balanced equation, 1 mole of \( \text{CH}_4 \) yields 2 moles of \( \text{H}_2\text{O} \).
- Therefore, 0.5 moles of \( \text{CH}_4 \) yields \( 0.5 \times 2 = 1.0 \text{ mole} \) of \( \text{H}_2\text{O} \).
- Mass of 1 mole of water = \( 18 \text{ g/mol} \). Mass formed = \( 1.0 \times 18 = 18 \text{ grams} \).
Why other options are incorrect:- 21, 19, 15 grams: These incorrect options result from failing to balance the equation correctly (forgetting that 1 \( \text{CH}_4 \) yields 2 \( \text{H}_2\text{O} \)) or miscalculating the molar mass of methane.
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