A mixture in which the mole ratio of `H_(2)` and `O_(2)` is `2:1` is used to prepare water by the reaction.
`2H_(2(g))+O_(2(g))rarr2H_(2)O_((g))`
The total pressure in the container is `0.8 atm` at `20^(@)C` before the reaction. Determine the final pressure at `120^(@)C` after reaction assuming `80%` yield of water.

Options

A. 1.787 atm
B. 0.878 atm
C. 0.787 atm
D. 1.878 atm

Explanation

To solve the problem step by step, we will follow the reaction and the conditions given in the question.

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction of hydrogen and oxygen to form water is:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \]

### Step 2: Define the initial moles of reactants
Let:
- The number of moles of \( H_2 \) be \( 2A \)
- The number of moles of \( O_2 \) be \( A \)

### Step 3: Calculate the initial total moles
The total initial moles before the reaction is:
\[ n_{initial} = 2A + A = 3A \]

### Step 4: Determine the yield of water
Given that the yield of water is 80%, we can express the moles of water produced:
- The theoretical moles of water produced from \( 2A \) moles of \( H_2 \) and \( A \) moles of \( O_2 \) is \( 2A \).
- The actual moles of water produced (80% yield) is:
\[ n_{H_2O} = 0.8 \times 2A = 1.6A \]

### Step 5: Calculate moles of reactants used
From the stoichiometry of the reaction:
- Moles of \( H_2 \) used = \( 0.8A \) (since \( 2 \text{ moles of } H_2 \) produce \( 2 \text{ moles of } H_2O \))
- Moles of \( O_2 \) used = \( 0.4A \) (since \( 1 \text{ mole of } O_2 \) produces \( 2 \text{ moles of } H_2O \))

### Step 6: Calculate remaining moles of reactants
After the reaction:
- Remaining moles of \( H_2 \):
\[ n_{H_2} = 2A - 0.8A = 1.2A \]
- Remaining moles of \( O_2 \):
\[ n_{O_2} = A - 0.4A = 0.6A \]

### Step 7: Calculate total moles after the reaction
Total moles after the reaction:
\[ n_{final} = n_{H_2} + n_{O_2} + n_{H_2O} \]
\[ n_{final} = 1.2A + 0.6A + 1.6A = 3.4A \]

### Step 8: Use the ideal gas law to find the final pressure
Using the ideal gas law \( PV = nRT \), we can find the pressure before and after the reaction.

**Initial conditions:**
- Pressure \( P_{initial} = 0.8 \, \text{atm} \)
- Temperature \( T_{initial} = 20^\circ C = 293 \, \text{K} \)
- Total moles \( n_{initial} = 3A \)

Using the ideal gas law:
\[ V = \frac{nRT}{P} \]
\[ V = \frac{3A \cdot R \cdot 293}{0.8} \]

**Final conditions:**
- Temperature \( T_{final} = 120^\circ C = 393 \, \text{K} \)
- Total moles \( n_{final} = 3.4A \)

Setting the volumes equal (since the volume is constant):
\[ \frac{3A \cdot R \cdot 293}{0.8} = \frac{3.4A \cdot R \cdot 393}{P_{final}} \]

### Step 9: Solve for \( P_{final} \)
Rearranging the equation:
\[ P_{final} = \frac{3.4A \cdot R \cdot 393 \cdot 0.8}{3A \cdot R \cdot 293} \]
\[ P_{final} = \frac{3.4 \cdot 393 \cdot 0.8}{3 \cdot 293} \]

Calculating:
\[ P_{final} = \frac{3.4 \cdot 393 \cdot 0.8}{879} \]
\[ P_{final} = 0.787 \, \text{atm} \]

### Final Answer
The final pressure at \( 120^\circ C \) after the reaction, assuming 80% yield of water, is:
\[ P_{final} = 0.787 \, \text{atm} \]

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