A 3 L gas mixture of propane `(C_(3)H_(8))` and butane `(C_(4)H_(10))` on complete combustion at 25 C produced 10 L `CO_(2.)` Assuming constant P and T conditions what was volume of butane present in initial mixture?

Options

A. `C_(3)H_(8),2LandC_(4)H_(10),1L`
B. `C_(3)H_(8),3LandC_(4)H_(10),0L`
C. `C_(3)H_(8),1.5L andC_(4)H_(10),1.5L`
D. `C_(3)H_(8),0LandC_(4)H_(10),3L`

Answer

To solve the problem step by step, we will analyze the combustion reactions of propane and butane, set up equations based on the information given, and solve for the volume of butane in the initial mixture.

### Step 1: Write the combustion reactions
1. **Combustion of Propane (C₃H₈)**:
\[ C₃H₈ + 5O₂ \rightarrow 3CO₂ + 4H₂O \]
From this reaction, we see that 1 volume of propane produces 3 volumes of carbon dioxide.

2. **Combustion of Butane (C₄H₁₀)**:
\[ C₄H₁₀ + \frac{13}{2}O₂ \rightarrow 4CO₂ + 5H₂O \]
Here, 1 volume of butane produces 4 volumes of carbon dioxide.

### Step 2: Set up the variables
Let:
- \( x \) = volume of propane in the mixture (in liters)
- \( 3 - x \) = volume of butane in the mixture (in liters)

### Step 3: Calculate the volume of CO₂ produced
From the combustion reactions:
- Volume of CO₂ from propane = \( 3x \) (since 1 volume of propane produces 3 volumes of CO₂)
- Volume of CO₂ from butane = \( 4(3 - x) \) (since 1 volume of butane produces 4 volumes of CO₂)

### Step 4: Set up the equation for total CO₂ produced
According to the problem, the total volume of CO₂ produced is 10 liters:
\[ 3x + 4(3 - x) = 10 \]

### Step 5: Simplify the equation
Expanding the equation:
\[ 3x + 12 - 4x = 10 \]
Combining like terms:
\[ -1x + 12 = 10 \]
Subtracting 12 from both sides:
\[ -x = -2 \]
Thus:
\[ x = 2 \]

### Step 6: Calculate the volume of butane
Now that we have \( x \), we can find the volume of butane:
\[ \text{Volume of butane} = 3 - x = 3 - 2 = 1 \text{ liter} \]

### Conclusion
The volume of butane present in the initial mixture is **1 liter**.

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