To determine the theoretical yield of water formed from the reaction of methane (CHâ‚„) with oxygen gas (Oâ‚‚), we'll follow these steps:
1. Write the balanced chemical equation.
2. Determine the limiting reactant.
3. Calculate the theoretical yield of water based on the limiting reactant.
### Step 1: Balanced Chemical Equation
The balanced chemical equation for the combustion of methane is:
\[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \]
### Step 2: Determine the Limiting Reactant
#### Calculate Moles of Each Reactant
**Moles of CHâ‚„:**
First, find the molar mass of CHâ‚„.
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol \(\times\) 4 = 4.04 g/mol
- Total molar mass of CHâ‚„ = 12.01 g/mol + 4.04 g/mol = 16.05 g/mol
\[ \text{Moles of CH}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{11.9 \text{ g}}{16.05 \text{ g/mol}} \approx 0.741 \text{ mol} \]
**Moles of Oâ‚‚:**
Molar mass of Oâ‚‚ = 32.00 g/mol (as each oxygen atom is 16.00 g/mol).
\[ \text{Moles of O}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{14.2 \text{ g}}{32.00 \text{ g/mol}} \approx 0.444 \text{ mol} \]
#### Determine the Limiting Reactant
From the balanced chemical equation, the stoichiometric ratio is:
\[ 1 \text{ mol CH}_4 : 2 \text{ mol O}_2 \]
Calculate how many moles of Oâ‚‚ are needed for the moles of CHâ‚„ we have:
\[ \text{Required moles of O}_2 = 0.741 \text{ mol CH}_4 \times 2 \text{ mol O}_2/\text{mol CH}_4 = 1.482 \text{ mol O}_2 \]
We have only 0.444 mol Oâ‚‚, which is less than 1.482 mol needed. Thus, Oâ‚‚ is the limiting reactant.
### Step 3: Calculate Theoretical Yield of Water
From the balanced equation, the mole ratio of Oâ‚‚ to Hâ‚‚O is:
\[ 2 \text{ mol O}_2 \rightarrow 2 \text{ mol H}_2\text{O} \]
Since Oâ‚‚ is the limiting reactant:
\[ \text{Moles of H}_2\text{O} = 0.444 \text{ mol O}_2 \times \frac{2 \text{ mol H}_2\text{O}}{2 \text{ mol O}_2} = 0.444 \text{ mol H}_2\text{O} \]
#### Calculate Mass of Hâ‚‚O
Molar mass of Hâ‚‚O = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
\[ \text{Mass of H}_2\text{O} = \text{moles} \times \text{molar mass} = 0.444 \text{ mol} \times 18.02 \text{ g/mol} \approx 8.00 \text{ g} \]
### Conclusion
The theoretical yield of water formed from the reaction of 11.9 g of methane and 14.2 g of oxygen gas is approximately 8.00 g.