A closed 10 L vessel contains 1 L water gas (1:1 `CO:H_2`) and 9 L air (20% `O_2` by volume) at STP. The contents of the vessel are ignited. The number of moles of `CO_2` in the vessel is closest to :

Options

A. 0.22
B. 0.022
C. 0.9
D. 3.6

Explanation

To solve the problem, we need to determine the number of moles of \( CO_2 \) produced when the contents of the closed vessel are ignited. The steps to find the solution are as follows:

### Step 1: Determine the composition of the gases in the vessel
The vessel contains:
- 1 L of water gas, which is a mixture of \( CO \) and \( H_2 \) in a 1:1 ratio.
- 9 L of air, which contains 20% \( O_2 \) by volume.

### Step 2: Calculate the volumes of \( CO \) and \( H_2 \)
Since the water gas is in a 1:1 ratio:
- Volume of \( CO = \frac{1}{2} \text{ L} \)
- Volume of \( H_2 = \frac{1}{2} \text{ L} \)

### Step 3: Calculate the volume of \( O_2 \) in the air
The air occupies 9 L, and \( O_2 \) constitutes 20% of this volume:
- Volume of \( O_2 = 0.20 \times 9 \text{ L} = 1.8 \text{ L} \)

### Step 4: Write the combustion reaction
The combustion of \( CO \) and \( H_2 \) can be represented as:
1. \( 2CO + O_2 \rightarrow 2CO_2 \)
2. \( 2H_2 + O_2 \rightarrow 2H_2O \)

### Step 5: Determine the limiting reactant
From the volumes:
- \( CO \): 0.5 L
- \( H_2 \): 0.5 L
- \( O_2 \): 1.8 L

Using the stoichiometry of the reactions:
- For \( 0.5 \text{ L of } CO \): \( 0.5 \text{ L CO} \) requires \( 0.25 \text{ L O}_2 \) (as \( 2CO \) requires \( 1O_2 \)).
- For \( 0.5 \text{ L of } H_2 \): \( 0.5 \text{ L H}_2 \) requires \( 0.25 \text{ L O}_2 \) (as \( 2H_2 \) requires \( 1O_2 \)).

Total \( O_2 \) required = \( 0.25 + 0.25 = 0.5 \text{ L} \).

Since we have 1.8 L of \( O_2 \), it is in excess, and \( CO \) and \( H_2 \) will be the limiting reactants.

### Step 6: Calculate the moles of \( CO_2 \) produced
From the combustion of \( CO \):
- \( 0.5 \text{ L CO} \) produces \( 0.5 \text{ L CO}_2 \).
Using the ideal gas law at STP (22.4 L/mol):
- Moles of \( CO_2 = \frac{0.5 \text{ L}}{22.4 \text{ L/mol}} \approx 0.02232 \text{ moles} \).

### Final Answer
The number of moles of \( CO_2 \) produced is approximately **0.02232 moles**.

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