Questions: Chemical Reactions Limiting reactants Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 0.16 g of methane is mixed with 0.205 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits. g × 10

Chemical Reactions Limiting reactants

Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 0.16 g of methane is mixed with 0.205 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits. g × 10

Transcript text: Chemical Reactions Limiting reactants Gaseous methane $\left(\mathrm{CH}_{4}\right)$ will react with gaseous oxygen $\left(\mathrm{O}_{2}\right)$ to produce gaseous carbon dioxide $\left(\mathrm{CO}_{2}\right)$ and gaseous water $\left(\mathrm{H}_{2} \mathrm{O}\right)$. Suppose 0.16 g of methane is mixed with 0.205 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits. $\square$ g $\times 10$

Solution

Solution Steps

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the reaction between methane ($\mathrm{CH}_4$) and oxygen ($\mathrm{O}_2$) is:

\[ \mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \]

Step 2: Calculate Moles of Reactants

First, calculate the moles of methane ($\mathrm{CH}_4$) and oxygen ($\mathrm{O}_2$) using their molar masses.

Molar mass of $\mathrm{CH}_4$ = 12.01 (C) + 4 \times 1.008 (H) = 16.04 \, \text{g/mol}
Moles of $\mathrm{CH}_4$ = $\frac{0.16 \, \text{g}}{16.04 \, \text{g/mol}} = 0.009975 \, \text{mol}$
Molar mass of $\mathrm{O}_2$ = 2 \times 16.00 = 32.00 \, \text{g/mol}
Moles of $\mathrm{O}_2$ = $\frac{0.205 \, \text{g}}{32.00 \, \text{g/mol}} = 0.00640625 \, \text{mol}$

Step 3: Determine the Limiting Reactant

According to the balanced equation, 1 mole of $\mathrm{CH}_4$ reacts with 2 moles of $\mathrm{O}_2$. Therefore, the moles of $\mathrm{O}_2$ required for 0.009975 moles of $\mathrm{CH}_4$ is:

\[ 0.009975 \, \text{mol} \times 2 = 0.01995 \, \text{mol} \]

Since we only have 0.00640625 moles of $\mathrm{O}_2$, $\mathrm{O}_2$ is the limiting reactant.

Step 4: Calculate Maximum Mass of Water Produced

From the balanced equation, 2 moles of $\mathrm{O}_2$ produce 2 moles of $\mathrm{H}_2\mathrm{O}$. Therefore, 0.00640625 moles of $\mathrm{O}_2$ will produce 0.00640625 moles of $\mathrm{H}_2\mathrm{O}$.

Molar mass of $\mathrm{H}_2\mathrm{O}$ = 2 \times 1.008 (H) + 16.00 (O) = 18.016 \, \text{g/mol}
Mass of $\mathrm{H}_2\mathrm{O}$ = $0.00640625 \, \text{mol} \times 18.016 \, \text{g/mol} = 0.1154 \, \text{g}$