Gaseous methane (CH4) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). What is the theoretical yield of water formed from the reaction of 11.9 g of methane and 14.2 g of oxygen gas?

To determine the theoretical yield of water formed from the reaction of methane (CH₄) with oxygen gas (O₂), 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 CH₄:**

First, find the molar mass of CH₄.
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol \(\times\) 4 = 4.04 g/mol
- Total molar mass of CH₄ = 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 O₂:**

Molar mass of O₂ = 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 O₂ are needed for the moles of CH₄ 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 O₂, which is less than 1.482 mol needed. Thus, O₂ is the limiting reactant.

### Step 3: Calculate Theoretical Yield of Water

From the balanced equation, the mole ratio of O₂ to H₂O is:

\[ 2 \text{ mol O}_2 \rightarrow 2 \text{ mol H}_2\text{O} \]

Since O₂ 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 H₂O

Molar mass of H₂O = 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.