A fuell cell uses `CH_(4)(g)` and forms `CO_(3)^(2-)` at the anode. It is used to power a car with 80 amp, for 0.96 hr. how many litres of `CH_(4)(g)` (at 1 atm, 273 K) would be required? (V_(m) = 22.4 L/mol) (F = 96500). Assume `100%` efficiency.

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the total charge passed through the fuel cell.
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I \) is the current in amperes (80 A)
- \( t \) is the time in seconds (0.96 hours)

First, convert hours to seconds:
\[ t = 0.96 \, \text{hr} \times 3600 \, \text{s/hr} = 3456 \, \text{s} \]

Now, calculate the total charge:
\[ Q = 80 \, \text{A} \times 3456 \, \text{s} = 276480 \, \text{C} \]

### Step 2: Calculate the number of moles of electrons transferred.
Using Faraday's law, the number of moles of electrons (n) can be calculated using the formula:
\[ n = \frac{Q}{F} \]
where \( F \) is Faraday's constant (96500 C/mol).

Now substitute the values:
\[ n = \frac{276480 \, \text{C}}{96500 \, \text{C/mol}} \approx 2.864 \, \text{mol} \]

### Step 3: Determine the number of moles of CH₄ consumed.
From the problem, we know that the n-factor for CH₄ (when it forms CO₃²⁻) is 8. This means that 8 moles of electrons are required to oxidize 1 mole of CH₄. Therefore, the number of moles of CH₄ (n_CH₄) can be calculated as:
\[ n_{CH₄} = \frac{n}{\text{n-factor}} = \frac{2.864 \, \text{mol}}{8} \approx 0.358 \, \text{mol} \]

### Step 4: Calculate the mass of CH₄ required.
The molar mass of CH₄ is approximately 16 g/mol. Thus, the mass of CH₄ can be calculated as:
\[ \text{mass}_{CH₄} = n_{CH₄} \times \text{molar mass}_{CH₄} = 0.358 \, \text{mol} \times 16 \, \text{g/mol} \approx 5.728 \, \text{g} \]

### Step 5: Convert mass of CH₄ to volume.
Using the molar volume at STP (22.4 L/mol), we can find the volume of CH₄:
\[ \text{volume}_{CH₄} = n_{CH₄} \times V_m = 0.358 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 8.020 \, \text{L} \]

### Final Answer:
The volume of CH₄ required is approximately **8.02 liters**.

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