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

Options

A. 11.2 L
B. 22.4 L
C. 2.24 L
D. 1.12 L

Answer

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

### Step 1: Write the balanced chemical equation
When carbon (C) burns in oxygen (O\(_2\)), it forms carbon dioxide (CO\(_2\)). The balanced chemical equation for this reaction is:
\[ C + O_2 \rightarrow CO_2 \]

### Step 2: Determine the molar mass of carbon
The atomic mass of carbon (C) is given as 12 g/mol.

### Step 3: Calculate the number of moles of carbon
To find the number of moles of carbon in 12 g, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles of C} = \frac{12 \, \text{g}}{12 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 4: Use the stoichiometry of the reaction
From the balanced equation, we see that 1 mole of carbon produces 1 mole of carbon dioxide (CO\(_2\)). Therefore, 1 mole of carbon will produce 1 mole of CO\(_2\).

### Step 5: Calculate the volume of \( CO_2 \) at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Thus, the volume of \( CO_2 \) produced from 1 mole of carbon is:
\[ \text{Volume of } CO_2 = 1 \, \text{mol} \times 22.4 \, \text{L/mol} = 22.4 \, \text{L} \]

### Final Answer
The volume of \( CO_2 \) liberated at STP when 12 g of carbon is burnt in excess oxygen is **22.4 liters**.

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