During discharging of lead-storage acid battery following reaction takes place :
`Pb(s) +PbO_(2)(s)+2H_(2)SO_(4)rarr 2PbSO_(4)(s)+2H_(2)O`
If `2.5` amp of current is drawn for 965 minutes, `H_(2)SO_(4)` consumed is :

Options

A. `0.75 ` mol
B. `3.00 ` mol
C. `1.50` mol
D. `4.50` mol

Answer

To solve the problem of how much sulfuric acid (H₂SO₄) is consumed during the discharging of a lead-storage acid battery when a current of 2.5 A is drawn for 965 minutes, we can follow these steps:

### Step 1: Convert Time to Seconds
First, we need to convert the time from minutes to seconds since the current is in amperes (A), which is coulombs per second.

\[ \text{Time in seconds} = 965 \text{ minutes} \times 60 \text{ seconds/minute} = 57900 \text{ seconds} \]

### Step 2: Calculate Total Charge (Q)
Using the formula \( Q = I \times t \), where \( I \) is the current in amperes and \( t \) is the time in seconds, we can calculate the total charge.

\[ Q = 2.5 \text{ A} \times 57900 \text{ s} = 144750 \text{ C} \]

### Step 3: Determine the Number of Moles of Electrons
Using Faraday's constant \( F \approx 96500 \text{ C/mol} \), we can find the number of moles of electrons (n) transferred.

\[ n = \frac{Q}{F} = \frac{144750 \text{ C}}{96500 \text{ C/mol}} \approx 1.5 \text{ mol of electrons} \]

### Step 4: Relate Moles of Electrons to Moles of H₂SO₄
From the balanced reaction, we can see that 2 moles of electrons are involved in the formation of 1 mole of H₂SO₄. Therefore, the number of moles of H₂SO₄ consumed can be calculated as follows:

\[ \text{Moles of H₂SO₄} = \frac{n}{2} = \frac{1.5}{2} = 0.75 \text{ mol} \]

### Step 5: Calculate the Mass of H₂SO₄ Consumed
Using the molar mass of H₂SO₄, which is approximately 98 g/mol, we can find the mass of H₂SO₄ consumed.

\[ \text{Mass of H₂SO₄} = \text{Moles} \times \text{Molar Mass} = 0.75 \text{ mol} \times 98 \text{ g/mol} = 73.5 \text{ g} \]

### Final Answer
The mass of H₂SO₄ consumed during the discharging process is approximately **73.5 grams**.

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