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: Understanding the Question:
The question asks for the mass of water produced from the complete combustion of $32\text{ g}$ of methane ($\text{CH}_4$) in the presence of excess oxygen.
This is a stoichiometry problem.

Step 2: Key Formula or Approach:
1. Write the balanced chemical equation for the combustion of methane.
2. Convert the given mass of methane to moles using the formula:
\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]
3. Use the stoichiometric ratio from the balanced equation to find the moles of water produced.
4. Convert the moles of water back to mass.

Step 3: Detailed Explanation:

• Step 3.1: Balanced Chemical Equation
The combustion of methane is represented as:
\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g) \]
From this equation, we see that $1\text{ mole}$ of $\text{CH}_4$ reacts to produce $2\text{ moles}$ of $\text{H}_2\text{O}$.

• Step 3.2: Calculate Moles of Methane
Given mass of $\text{CH}_4 = 32\text{ g}$
Molar mass of $\text{CH}_4 = 16\text{ g/mol}$
\[ \text{Moles of }\text{CH}_4 = \frac{32\text{ g}}{16\text{ g/mol}} = 2\text{ moles} \]

• Step 3.3: Calculate Moles of Water Produced
Using the stoichiometric coefficient ratio ($\text{CH}_4 : \text{H}_2\text{O} = 1 : 2$):
\[ \text{Moles of }\text{H}_2\text{O} = 2 \times \text{Moles of }\text{CH}_4 = 2 \times 2 = 4\text{ moles} \]

• Step 3.4: Calculate Mass of Water Produced
Molar mass of $\text{H}_2\text{O} = 18\text{ g/mol}$
\[ \text{Mass of }\text{H}_2\text{O} = \text{Moles of }\text{H}_2\text{O} \times \text{Molar Mass of }\text{H}_2\text{O} \]
\[ \text{Mass of }\text{H}_2\text{O} = 4\text{ moles} \times 18\text{ g/mol} = 72\text{ g} \]


Step 4: Final Answer:
The total mass of water produced is $72\text{ g}$.