Question

Gaseous methane (CH4) reacts with gaseous oxygen gas (O2)
to produce gaseous carbon dioxide (CO2) and gaseous water (H2O)
. If 0.122g of water is produced from the reaction of 0.16g
of methane and 0.25g of oxygen gas, calculate the percent yield of water.

Answer 1

$$\mathrm{CH_4} + 2\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\mathrm{H_2O}$$ $$\mathrm{m(CH_4)} = 0.16\ \mathrm{g}$$ $$\mathrm{m(O_2)} = 0.25\ \mathrm{g}$$ $$\mathrm{M(CH_4)} = 12 + 1.4 = 16\ \mathrm{g/mol}$$ $$\mathrm{M(O_2)} = 16.2 = 32\ \mathrm{g/mol}$$ $$\mathrm{n(CH_4)} = \frac{m}{M} = \frac{0.16\ \mathrm{g}}{16\ \frac{\mathrm{g}}{\mathrm{mol}}} = 0.01\ \mathrm{mol}$$ $$\mathrm{n(O_2)} = \frac{m}{M} = \frac{0.25\ \mathrm{g}}{32\ \frac{\mathrm{g}}{\mathrm{mol}}} = 0.0078\ \mathrm{mol}$$

$\mathrm{O_2}$ is the limiting reactant/

$$\mathrm{n(H_2O)} = 0.0078\ \mathrm{mol}$$ $$\mathrm{m(H_2O)} = \mathrm{n\cdot M} = 0.0078\ \mathrm{mol\cdot 18g/mol} = 0.14\ \mathrm{g}$$

Percent yield = $\frac{\text{Actual mass of product}}{\text{Predicted mass of product}} \cdot 100\%$

Percent yield of water = $\frac{0.122\ \mathrm{g}}{0.14\ \mathrm{g}} = 87.14\%$

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