what volume of `CO_2` will be liberated at STP if 12 g of carbon is burnt in excess of oxygen ?

Options

A. 11.35 L
B. 22.7 L
C. 2.27 L
D. 1.135 L

Answer

To solve the problem of how much volume of \( CO_2 \) will be liberated at STP when 12 g of carbon is burnt in excess of oxygen, we can follow these steps:

### Step 1: Write the balanced chemical equation
The combustion of carbon in oxygen can be represented by the following balanced equation:
\[ C + O_2 \rightarrow CO_2 \]
This shows that one mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide.

### Step 2: Determine the molar mass of carbon
The molar mass of carbon (C) is approximately 12 g/mol. This means that 12 grams of carbon is equivalent to:
\[ \text{Number of moles of carbon} = \frac{\text{mass of carbon}}{\text{molar mass of carbon}} = \frac{12 \text{ g}}{12 \text{ g/mol}} = 1 \text{ mol} \]

### Step 3: Relate moles of carbon to moles of \( CO_2 \)
From the balanced equation, we see that 1 mole of carbon produces 1 mole of \( CO_2 \). Therefore, burning 1 mole of carbon will produce:
\[ 1 \text{ mol of } CO_2 \]

### Step 4: Calculate the volume of \( CO_2 \) at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies a volume of 22.4 liters. Therefore, the volume of \( CO_2 \) produced from 1 mole is:
\[ \text{Volume of } CO_2 = 1 \text{ mol} \times 22.4 \text{ L/mol} = 22.4 \text{ L} \]

### Conclusion
Thus, the volume of \( CO_2 \) liberated at STP when 12 g of carbon is burnt in excess of oxygen is:
\[ \text{Volume of } CO_2 = 22.4 \text{ L} \]