Volume at NTP of oxygen required to completely burn 1 kg of coal (100% carbon)

Options

A. 22400 L
B. `22.4 xx 10^(3) L`
C. `1.86 xx 10^(3) L`
D. 1000L

Answer

To solve the problem of finding the volume at NTP of oxygen required to completely burn 1 kg of coal (100% carbon), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the mass of carbon:**
The mass of coal given is 1 kg, which is equivalent to 1000 grams.

\[ \text{Mass of carbon} = 1000 \text{ g} \]

2. **Calculate the number of moles of carbon:**
The molar mass of carbon (C) is approximately 12 g/mol. To find the number of moles of carbon, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1000 \text{ g}}{12 \text{ g/mol}} \approx 83.34 \text{ moles} \]

3. **Write the balanced chemical equation for combustion:**
The combustion of carbon can be represented by the following equation:

\[ C + O_2 \rightarrow CO_2 \]

From this equation, we see that 1 mole of carbon reacts with 1 mole of oxygen.

4. **Determine the moles of oxygen required:**
Since 1 mole of carbon requires 1 mole of oxygen, the moles of oxygen required for 83.34 moles of carbon will also be 83.34 moles.

\[ \text{Moles of } O_2 = 83.34 \text{ moles} \]

5. **Calculate the volume of oxygen at NTP:**
At Normal Temperature and Pressure (NTP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of oxygen required can be calculated as follows:

\[ \text{Volume of } O_2 = \text{moles of } O_2 \times 22.4 \text{ L/mol} = 83.34 \text{ moles} \times 22.4 \text{ L/mol} \approx 1866.336 \text{ L} \]

Rounding this to two significant figures gives us approximately:

\[ \text{Volume of } O_2 \approx 1.86 \times 10^3 \text{ L} \]

6. **Conclusion:**
The volume of oxygen required to completely burn 1 kg of coal at NTP is approximately \( 1.86 \times 10^3 \) liters.

### Final Answer:
The volume at NTP of oxygen required to completely burn 1 kg of coal (100% carbon) is \( 1.86 \times 10^3 \) liters.

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