Chief ore of `Zn` is `ZnS`. The ore is concentrated by froth flotation process and then heated in air to convert `ZnS` to `ZnO`. ltntgt `2 ZnS + 3O_(2) rarr 2 ZnO + 2 SO_(2)`…(i)
`ZnO`, thus formed is treated with dilute `H_(2) SO_(4)`.
`ZnO + H_(2) SO_(4) rarr ZnSO_(4) + H_(2) O` ....(ii)
On electrolysis of `ZnSO_(4(aq)), Zn` metal is produced.
`2 ZnSO_(4) + H_(2) O rarr 2 Zn + 2H_(2) SO_(4) + O_(2)`....(iii)
What volume of `98 % H_(2) SO_(4)` (by weight, density `= 1.8 g//mL`) is required in step `(ii)` ?

Options

A. 120 L
B. 129 L
C. 1.40 L
D. 150 L

Explanation

To solve the problem step by step, we will follow the reactions and calculations outlined in the video transcript.

### Step 1: Understanding the Reaction
The first reaction involves the conversion of zinc sulfide (ZnS) to zinc oxide (ZnO) through heating in the presence of oxygen:
\[ 2 \text{ZnS} + 3 \text{O}_2 \rightarrow 2 \text{ZnO} + 2 \text{SO}_2 \]

### Step 2: Moles of ZnS
Given that we have 225 kg of ZnS, we need to convert this mass into moles. The molar mass of ZnS is approximately:
- Zn: 65 g/mol
- S: 32 g/mol
- Molar mass of ZnS = 65 + 32 = 97 g/mol

Now, converting 225 kg to grams:
\[ 225 \text{ kg} = 225 \times 10^3 \text{ g} \]

Calculating the number of moles of ZnS:
\[ \text{Moles of ZnS} = \frac{225 \times 10^3 \text{ g}}{97 \text{ g/mol}} \approx 2320.62 \text{ moles} \]

### Step 3: Moles of ZnO Produced
From the stoichiometry of the reaction, 2 moles of ZnS produce 2 moles of ZnO. Therefore, the moles of ZnO produced will be equal to the moles of ZnS:
\[ \text{Moles of ZnO} = 2320.62 \text{ moles} \]

### Step 4: Reaction with H2SO4
The next step involves treating ZnO with dilute sulfuric acid (H2SO4):
\[ \text{ZnO} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2\text{O} \]

From the balanced equation, 1 mole of ZnO reacts with 1 mole of H2SO4. Therefore, the moles of H2SO4 required will also be:
\[ \text{Moles of H2SO4} = 2320.62 \text{ moles} \]

### Step 5: Mass of H2SO4 Required
The molar mass of H2SO4 is approximately:
- H: 1 g/mol (2 H atoms)
- S: 32 g/mol
- O: 16 g/mol (4 O atoms)
- Molar mass of H2SO4 = 2 + 32 + 64 = 98 g/mol

Calculating the mass of H2SO4 required:
\[ \text{Mass of H2SO4} = \text{Moles} \times \text{Molar Mass} \]
\[ \text{Mass of H2SO4} = 2320.62 \text{ moles} \times 98 \text{ g/mol} \approx 227,000 \text{ g} \]

### Step 6: Volume of H2SO4 Required
To find the volume of 98% H2SO4 solution required, we need to consider the density of the solution (1.8 g/mL) and the fact that it is 98% by weight.

First, calculate the mass of pure H2SO4 in the solution:
\[ \text{Mass of pure H2SO4} = 0.98 \times \text{Mass of H2SO4 solution} \]

Let \( x \) be the mass of the solution:
\[ 0.98x = 227,000 \text{ g} \]
\[ x = \frac{227,000}{0.98} \approx 231,632.65 \text{ g} \]

Now, convert this mass into volume using the density:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
\[ \text{Volume} = \frac{231,632.65 \text{ g}}{1.8 \text{ g/mL}} \approx 128,595.36 \text{ mL} \]
\[ \text{Volume} \approx 128.60 \text{ L} \]

### Final Answer
The volume of 98% H2SO4 required in step (ii) is approximately **128.60 L**.

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