Question

1 balanced the equation for this reaction, and answer the questions.
$\boldsymbol{2 c(s) + 2 h_2o(g)
ightarrow _ ch_4(g) + _ co_2(g)}$
a) how many grams of methane, $\text{ch}_4$, can be produced from 235.0 g carbon reacted in an excess of water?
b) if 126.0 g $\text{ch}_4$ are actually collected in the lab, what is the percent yield?

Explanation:

Step1: Balance the chemical equation

First, balance C, H, then O atoms:

$$2\text{C}(s) + 2\text{H}_2\text{O}(g) ightarrow \text{CH}_4(g) + \text{CO}_2(g)$$

Step2: Calculate moles of C

Molar mass of C = $12.01\ \text{g/mol}$
$$n_{\text{C}} = \frac{235.0\ \text{g}}{12.01\ \text{g/mol}} \approx 19.57\ \text{mol}$$

Step3: Relate moles of C to CH₄

From balanced equation: 2 mol C → 1 mol CH₄
$$n_{\text{CH}_4} = \frac{1}{2} \times n_{\text{C}} = \frac{1}{2} \times 19.57\ \text{mol} = 9.785\ \text{mol}$$

Step4: Find theoretical mass of CH₄

Molar mass of CH₄ = $16.04\ \text{g/mol}$
$$m_{\text{theoretical}} = 9.785\ \text{mol} \times 16.04\ \text{g/mol} \approx 156.9\ \text{g}$$

Step5: Calculate percent yield

Percent yield = $\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$
$$\text{Percent Yield} = \frac{126.0\ \text{g}}{156.9\ \text{g}} \times 100\% \approx 80.3\%$$

Answer:

Balanced equation: $2\text{C}(s) + 2\text{H}_2\text{O}(g)
ightarrow \text{CH}_4(g) + \text{CO}_2(g)$
a) Theoretical mass of CH₄: $\approx 156.9\ \text{g}$
b) Percent yield: $\approx 80.3\%$