Ethanol burns in excess oxygen to form `CO_2(g)` and `H_2O(g)` according to this balanced equation.
`C_2H_5OH(g)+3O_2(g) to2CO_2(g)+3H_2O(g)`
What value is closest to the volume of `CO_2(g)` , measured at 200 K and 1 atm produced from the combustion of 0.25 mol of `C_2H_5OH(g)`?

Options

A. 5 L
B. 8 L
C. 10 L
D. 15 L

Answer

To determine the volume of carbon dioxide (CO₂) produced from the combustion of 0.25 mol of ethanol (C₂H₅OH), we can follow these steps:

### Step 1: Analyze the Balanced Chemical Equation
The balanced equation for the combustion of ethanol is:
\[ \text{C}_2\text{H}_5\text{OH} (g) + 3\text{O}_2 (g) \rightarrow 2\text{CO}_2 (g) + 3\text{H}_2\text{O} (g) \]

From the equation, we can see that 1 mole of ethanol produces 2 moles of carbon dioxide.

### Step 2: Calculate Moles of CO₂ Produced
Given that we have 0.25 moles of ethanol, we can calculate the moles of CO₂ produced using the stoichiometry of the reaction:
\[ \text{Moles of CO}_2 = 0.25 \, \text{mol C}_2\text{H}_5\text{OH} \times \frac{2 \, \text{mol CO}_2}{1 \, \text{mol C}_2\text{H}_5\text{OH}} = 0.50 \, \text{mol CO}_2 \]

### Step 3: Use the Ideal Gas Law to Calculate Volume
We can use the Ideal Gas Law, which is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

Given:
- \( P = 1 \, \text{atm} \)
- \( n = 0.50 \, \text{mol} \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 200 \, \text{K} \)

We can rearrange the Ideal Gas Law to solve for volume \( V \):
\[ V = \frac{nRT}{P} \]

### Step 4: Substitute the Values into the Equation
Substituting the known values into the equation:
\[ V = \frac{0.50 \, \text{mol} \times 0.0821 \, \text{L·atm/(K·mol)} \times 200 \, \text{K}}{1 \, \text{atm}} \]

### Step 5: Calculate the Volume
Calculating the volume:
\[ V = \frac{0.50 \times 0.0821 \times 200}{1} = 8.21 \, \text{L} \]

### Conclusion
The volume of CO₂ produced from the combustion of 0.25 mol of ethanol at 200 K and 1 atm is approximately 8.21 liters.

### Final Answer
The closest value to the volume of CO₂ produced is **8.2 liters**.

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