A lead storage battery has been used for one month (30 days) at the rate of one hour per day by drawing a constant current of 2 amperes. `H_(2)SO_(4)` consumed by the battery is:-

Options

A. 1.12 mole
B. 2.24 mole
C. 3.36 mole
D. 4.48 mole

Explanation

To solve the problem of how much \( H_2SO_4 \) is consumed by a lead storage battery after being used for one month at a constant current of 2 amperes, we will follow these steps:

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

Since the battery is used for 1 hour each day for 30 days, we first convert the time into seconds:
\[ t = 1 \text{ hour/day} \times 30 \text{ days} = 30 \text{ hours} \]
Now, converting hours into seconds:
\[ t = 30 \text{ hours} \times 3600 \text{ seconds/hour} = 108000 \text{ seconds} \]

Now, substituting the values into the charge formula:
\[ Q = 2 \text{ A} \times 108000 \text{ s} = 216000 \text{ C} \]

### Step 2: Determine the number of moles of \( H_2SO_4 \) consumed
From the balanced chemical equation for the lead storage battery, we know that 2 moles of \( H_2SO_4 \) are consumed for every 2 moles of electrons transferred. The Faraday constant (the charge of one mole of electrons) is approximately \( 96500 \text{ C/mol} \).

Using the relationship:
\[ \text{Moles of electrons} = \frac{Q}{F} \]
where \( F \) is the Faraday constant.

Calculating the moles of electrons:
\[ \text{Moles of electrons} = \frac{216000 \text{ C}}{96500 \text{ C/mol}} \approx 2.24 \text{ mol} \]

Since 2 moles of \( H_2SO_4 \) are consumed per 2 moles of electrons, the moles of \( H_2SO_4 \) consumed will be equal to the moles of electrons:
\[ \text{Moles of } H_2SO_4 = 2.24 \text{ mol} \]

### Conclusion
The amount of \( H_2SO_4 \) consumed by the battery after one month of usage is approximately **2.24 moles**.

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