The total pressure of a mixture of oxygen and hydrogen is 1.0 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.40 atm when measured at the same values of T and V as the original mixture. What was the composition of the original mixture in mole fraction ?

Options

A. `x_(o_(2))=0.2,x_(H_(2))=0.8`
B. `x_(o_(2))=0.4,x_(H_(2))=0.6`
C. `x_(o_(2))=0.6,x_(H_(2))=0.4`
D. `x_(o_(2))=0.8,x_(H_(2))=0.2`

Explanation

To solve the problem, we need to determine the composition of the original mixture of hydrogen (H₂) and oxygen (O₂) in terms of mole fractions. We know the total pressure of the mixture before ignition and the pressure of the remaining gas after the reaction. Here’s a step-by-step solution:

### Step 1: Understand the Reaction
When hydrogen and oxygen react, they form water:
\[ 2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O} \]
From this reaction, we see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

### Step 2: Identify Given Information
- Total pressure of the mixture (P_total) = 1.0 atm
- Pressure of remaining pure hydrogen after ignition (P_H2) = 0.40 atm

### Step 3: Calculate Pressure Consumed
The pressure consumed during the reaction can be calculated as:
\[ P_{\text{consumed}} = P_{\text{total}} - P_{\text{H}_2} = 1.0 \, \text{atm} - 0.40 \, \text{atm} = 0.60 \, \text{atm} \]

### Step 4: Relate Consumed Pressure to Moles
From the stoichiometry of the reaction, we know that 2 moles of H₂ react with 1 mole of O₂. Therefore, the pressure consumed can be divided into contributions from H₂ and O₂.

Let:
- \( P_{\text{H}_2} = x \) atm (initial pressure of hydrogen)
- \( P_{\text{O}_2} = y \) atm (initial pressure of oxygen)

From the total pressure:
\[ x + y = 1.0 \, \text{atm} \]

### Step 5: Use Stoichiometry to Relate Pressures
From the stoichiometry of the reaction:
- For every 2 atm of H₂ consumed, 1 atm of O₂ is consumed.
- The total pressure consumed (0.60 atm) can be expressed as:
\[ \frac{2}{3} P_{\text{H}_2} + \frac{1}{3} P_{\text{O}_2} = 0.60 \, \text{atm} \]

### Step 6: Solve the System of Equations
We have two equations:
1. \( x + y = 1.0 \)
2. \( \frac{2}{3} x + \frac{1}{3} y = 0.60 \)

Substituting \( y = 1.0 - x \) into the second equation:
\[ \frac{2}{3} x + \frac{1}{3} (1.0 - x) = 0.60 \]
\[ \frac{2}{3} x + \frac{1}{3} - \frac{1}{3} x = 0.60 \]
\[ \frac{1}{3} x + \frac{1}{3} = 0.60 \]
\[ \frac{1}{3} x = 0.60 - \frac{1}{3} \]
\[ \frac{1}{3} x = 0.60 - 0.33 \]
\[ \frac{1}{3} x = 0.27 \]
\[ x = 0.81 \, \text{atm} \]

Now substituting back to find \( y \):
\[ y = 1.0 - 0.81 = 0.19 \, \text{atm} \]

### Step 7: Calculate Mole Fractions
Now we can calculate the mole fractions:
- Mole fraction of H₂ (\( X_{H2} \)):
\[ X_{H2} = \frac{P_{H2}}{P_{total}} = \frac{0.81}{1.0} = 0.81 \]

- Mole fraction of O₂ (\( X_{O2} \)):
\[ X_{O2} = \frac{P_{O2}}{P_{total}} = \frac{0.19}{1.0} = 0.19 \]

### Final Answer
The composition of the original mixture in mole fractions is:
- Mole fraction of H₂ = 0.81
- Mole fraction of O₂ = 0.19