Calculate the volume of `O_(2)` and volume of air needed for combustion of `1 kg` carbon at `STP`.

Options

A. 1866.6, 8888.5
B. 1866.6, 8433.35
C. 1866.6, 7233.35
D. 3666.6, 9433.35

Explanation

To solve the problem of calculating the volume of \( O_2 \) and the volume of air needed for the combustion of 1 kg of carbon at STP, we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of carbon.
The balanced equation for the combustion of carbon (C) is:
\[ C + O_2 \rightarrow CO_2 \]
This indicates that 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide.

### Step 2: Calculate the number of moles of carbon in 1 kg.
To find the number of moles of carbon, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of carbon (C) is 12 g/mol. Therefore, for 1 kg (1000 g) of carbon:
\[ \text{Number of moles of C} = \frac{1000 \, \text{g}}{12 \, \text{g/mol}} = 83.33 \, \text{moles} \]

### Step 3: Determine the number of moles of \( O_2 \) required.
From the balanced equation, we see that 1 mole of carbon requires 1 mole of \( O_2 \). Thus, the number of moles of \( O_2 \) required is also:
\[ \text{Number of moles of } O_2 = 83.33 \, \text{moles} \]

### Step 4: Calculate the volume of \( O_2 \) at STP.
At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.41 liters. Therefore, the volume of \( O_2 \) required can be calculated as:
\[ \text{Volume of } O_2 = \text{Number of moles of } O_2 \times 22.41 \, \text{L/mol} \]
\[ \text{Volume of } O_2 = 83.33 \, \text{moles} \times 22.41 \, \text{L/mol} = 1866.6 \, \text{liters} \]

### Step 5: Calculate the volume of air needed for combustion.
Air is composed of approximately 21% oxygen. To find the total volume of air needed to provide the required volume of \( O_2 \), we can set up the following proportion:
\[ \frac{1866.6 \, \text{L}}{x} = \frac{21}{100} \]
Where \( x \) is the total volume of air needed. Rearranging gives:
\[ x = \frac{1866.6 \, \text{L} \times 100}{21} \approx 8888.57 \, \text{liters} \]

### Final Answers:
- Volume of \( O_2 \) required: **1866.6 liters**
- Volume of air required: **8888.57 liters**