When burnt in air, 14.0 g mixture of carbon and sulphur gives a mixture of `CO_(2)` and `SO_(2)` in the volume ratio of 2 : 1, volumes being measured at the same conditions of temperature and pressure. Moles of carbon in the mixture is

Options

A. 0.25
B. 0.4
C. 0.5
D. 0.75

Explanation

To find the moles of carbon in the mixture, we can follow these steps:

### Step 1: Define the Variables
Let the weight of carbon in the mixture be \( x \) grams. Therefore, the weight of sulfur in the mixture will be \( 14 - x \) grams since the total weight of the mixture is 14 grams.

### Step 2: Write the Chemical Reactions
When carbon burns in air, it produces carbon dioxide (\( CO_2 \)):
\[ C + O_2 \rightarrow CO_2 \]
When sulfur burns in air, it produces sulfur dioxide (\( SO_2 \)):
\[ S + O_2 \rightarrow SO_2 \]

### Step 3: Determine Moles of Products
From the problem, we know that the volume ratio of \( CO_2 \) to \( SO_2 \) is \( 2:1 \). Since volumes are measured under the same conditions of temperature and pressure, we can relate the volumes to the number of moles:
\[ \text{Moles of } CO_2 : \text{Moles of } SO_2 = 2:1 \]

Let the moles of \( CO_2 \) be \( n \). Then, the moles of \( SO_2 \) will be \( \frac{n}{2} \).

### Step 4: Relate Moles to Mass
Using the molar masses:
- Molar mass of carbon (C) = 12 g/mol
- Molar mass of sulfur (S) = 32 g/mol

The number of moles of carbon can be expressed as:
\[ \text{Moles of } C = \frac{x}{12} \]
The number of moles of sulfur can be expressed as:
\[ \text{Moles of } S = \frac{14 - x}{32} \]

### Step 5: Set Up the Equation
From the stoichiometry of the reaction, we know:
\[ \text{Moles of } C = n \quad \text{and} \quad \text{Moles of } S = \frac{n}{2} \]
Thus, we can write:
\[ \frac{x}{12} = n \quad \text{and} \quad \frac{14 - x}{32} = \frac{n}{2} \]

### Step 6: Substitute and Solve
Substituting \( n \) from the first equation into the second:
\[ \frac{14 - x}{32} = \frac{1}{2} \cdot \frac{x}{12} \]
Cross-multiplying gives:
\[ 2(14 - x) = \frac{x \cdot 32}{12} \]
Simplifying:
\[ 28 - 2x = \frac{32x}{12} \]
Multiplying through by 12 to eliminate the fraction:
\[ 336 - 24x = 32x \]
Combining like terms:
\[ 336 = 56x \]
Solving for \( x \):
\[ x = \frac{336}{56} = 6 \text{ grams} \]

### Step 7: Calculate Moles of Carbon
Now that we have the weight of carbon, we can calculate the moles:
\[ \text{Moles of } C = \frac{6}{12} = 0.5 \text{ moles} \]

### Final Answer
The moles of carbon in the mixture is \( 0.5 \) moles.

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