By the reaction of carbon and oxygen, a mixture of `CO` and `CO_(2)` is obtained. What is the composition of the mixture by mass obtained when `20` grams of `O_(2)` reacts with `12` grams of carbon ?

Explanation

To solve the problem of determining the composition by mass of the mixture of carbon monoxide (CO) and carbon dioxide (CO₂) produced when 20 grams of O₂ reacts with 12 grams of carbon, we can follow these steps:

### Step 1: Calculate the moles of carbon and oxygen.
- **Molar mass of carbon (C)** = 12 g/mol
- **Molar mass of oxygen (O₂)** = 32 g/mol

**Moles of carbon (C)**:
\[ \text{Moles of C} = \frac{\text{mass of C}}{\text{molar mass of C}} = \frac{12 \text{ g}}{12 \text{ g/mol}} = 1 \text{ mol} \]

**Moles of oxygen (O₂)**:
\[ \text{Moles of O₂} = \frac{\text{mass of O₂}}{\text{molar mass of O₂}} = \frac{20 \text{ g}}{32 \text{ g/mol}} = 0.625 \text{ mol} \]

### Step 2: Write the balanced chemical equations.
The reactions that occur are:
1. \( 2C + O_2 \rightarrow 2CO \)
2. \( C + O_2 \rightarrow CO_2 \)

### Step 3: Determine the limiting reactant.
From the balanced equations:
- For every 1 mole of O₂, 2 moles of C are required to produce CO.
- For every 1 mole of O₂, 1 mole of C is required to produce CO₂.

Let \( x \) be the moles of CO₂ produced. Then, the moles of CO produced will be \( 1 - x \) (since all carbon is used).

From the reaction stoichiometry:
- For CO₂: \( x \) moles of C and O₂ are used.
- For CO: \( 1 - x \) moles of C and \( 0.5(1 - x) \) moles of O₂ are used.

The total moles of O₂ used can be expressed as:
\[ x + 0.5(1 - x) = 0.625 \]

### Step 4: Solve for \( x \).
Expanding the equation:
\[ x + 0.5 - 0.5x = 0.625 \]
\[ 0.5x + 0.5 = 0.625 \]
\[ 0.5x = 0.125 \]
\[ x = 0.25 \]

### Step 5: Calculate moles of CO produced.
Moles of CO produced:
\[ 1 - x = 1 - 0.25 = 0.75 \text{ mol} \]

### Step 6: Calculate the mass of CO and CO₂ produced.
**Mass of CO**:
\[ \text{Molar mass of CO} = 12 + 16 = 28 \text{ g/mol} \]
\[ \text{Mass of CO} = \text{moles of CO} \times \text{molar mass of CO} = 0.75 \text{ mol} \times 28 \text{ g/mol} = 21 \text{ g} \]

**Mass of CO₂**:
\[ \text{Molar mass of CO₂} = 12 + 2 \times 16 = 44 \text{ g/mol} \]
\[ \text{Mass of CO₂} = \text{moles of CO₂} \times \text{molar mass of CO₂} = 0.25 \text{ mol} \times 44 \text{ g/mol} = 11 \text{ g} \]

### Step 7: Calculate the composition by mass.
Total mass of the mixture:
\[ \text{Total mass} = \text{mass of CO} + \text{mass of CO₂} = 21 \text{ g} + 11 \text{ g} = 32 \text{ g} \]

Composition by mass:
- Mass percentage of CO:
\[ \text{Mass \% of CO} = \left( \frac{21 \text{ g}}{32 \text{ g}} \right) \times 100 = 65.625\% \]
- Mass percentage of CO₂:
\[ \text{Mass \% of CO₂} = \left( \frac{11 \text{ g}}{32 \text{ g}} \right) \times 100 = 34.375\% \]

### Final Answer:
The composition of the mixture by mass is approximately:
- CO: 65.63%
- CO₂: 34.38%