Calculate the weight of carbon which is burnt with excess of oxygen to form` 22.4 L` of` CO_2` at NTP

Options

A. 4 g
B. 6 g
C. 12 g
D. 24 g

Answer

To solve the problem of calculating the weight of carbon that is burnt with excess oxygen to form 22.4 L of CO₂ at NTP, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**:
The combustion of carbon in oxygen can be represented by the following balanced chemical equation:
\[ C + O_2 \rightarrow CO_2 \]
This means that 1 mole of carbon reacts with oxygen to produce 1 mole of carbon dioxide.

2. **Convert Volume of CO₂ to Moles**:
At Normal Temperature and Pressure (NTP), 1 mole of any ideal gas occupies 22.4 liters. Therefore, to find the number of moles of CO₂ produced from 22.4 L, we use the formula:
\[ \text{Number of moles of } CO_2 = \frac{\text{Volume of } CO_2}{\text{Molar volume at NTP}} = \frac{22.4 \, \text{L}}{22.4 \, \text{L/mol}} = 1 \, \text{mol} \]

3. **Determine Moles of Carbon Required**:
From the balanced equation, we see that 1 mole of carbon produces 1 mole of CO₂. Therefore, if we have 1 mole of CO₂, we also need 1 mole of carbon.

4. **Calculate the Mass of Carbon**:
The molar mass of carbon (C) is approximately 12 g/mol. Thus, the mass of 1 mole of carbon is:
\[ \text{Mass of carbon} = \text{Number of moles} \times \text{Molar mass} = 1 \, \text{mol} \times 12 \, \text{g/mol} = 12 \, \text{g} \]

5. **Conclusion**:
The weight of carbon that is burnt to produce 22.4 L of CO₂ at NTP is 12 grams.

### Final Answer:
The weight of carbon burnt is **12 grams**.

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