Question

If $32\text{ g}$ of methane reacts completely with oxygen, how much water is produced? ($\text{CH}_4 = 16\text{ g/mol}$, $\text{H}_2\text{O} = 18\text{ g/mol}$)

• A. $36\text{ g}$
• B. $54\text{ g}$
• C. $72\text{ g}$
• D. $18\text{ g}$

Hint:

We can also solve this quickly using mass-mass ratios:
$16\text{ g}$ of methane (1 mole) gives $36\text{ g}$ of water (2 moles).
Doubling the reactant mass to $32\text{ g}$ simply doubles the product mass to $72\text{ g}$.

Answer 1

The Correct Option is C

Step 1: Write the balanced combustion equation.
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]
One mole of methane produces two moles of water on complete combustion.
Step 2: Convert the given mass of methane to moles.
\[ \text{Moles of } \text{CH}_4 = \frac{32 \text{ g}}{16 \text{ g/mol}} = 2 \text{ mol} \]
Step 3: Use the mole ratio to find water produced.
\[ \text{Moles of } \text{H}_2\text{O} = 2 \times 2 = 4 \text{ mol} \]
\[ \text{Mass of } \text{H}_2\text{O} = 4 \times 18 = 72 \text{ g} \]
\[ \boxed{72\text{ g}} \]