Phospheric acid `(H_(3)PO_(4))` perpared in two step process .
(1) `P_(4)+5O_(2) to P_(4)O_(10)" "(2) P_(4)O_(10) + 6H_(2)O to 4H_(3)PO_(4)`
Well allow 62 g of phosphrous to react with exces oxygen which from `P_(4)O_(10)` in `85%` yield . In the sep (2) reaction `90%` yield of `H_3PO_(4)` is obtained . Mass of `H_(3) PO_(4)` produced is :

Options

A. `37.485 g`
B. `149. 949 g`
C. `125. 47 g `
D. `564.48 g`

Answer

To solve the problem step-by-step, we will follow the two-step reaction process given in the question.

### Step 1: Calculate Moles of P4
1. **Given Data**:
- Mass of phosphorus (P4) = 62 g
- Molar mass of phosphorus (P) = 31 g/mol
- Molar mass of P4 = 4 × 31 = 124 g/mol

2. **Calculate Moles of P4**:
\[ \text{Moles of } P4 = \frac{\text{mass of } P4}{\text{molar mass of } P4} = \frac{62 \, \text{g}}{124 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 2: Calculate Moles of P4O10 Produced
3. **Reaction**:
\[ P4 + 5O2 \rightarrow P4O10 \]
- From the stoichiometry of the reaction, 1 mole of P4 produces 1 mole of P4O10.

4. **Calculate Moles of P4O10**:
\[ \text{Moles of } P4O10 = \text{Moles of } P4 = 0.5 \, \text{mol} \]

5. **Considering the Yield**:
- Given yield of P4O10 = 85%
\[ \text{Moles obtained of } P4O10 = \text{Moles calculated} \times \frac{\text{yield}}{100} = 0.5 \times \frac{85}{100} = 0.425 \, \text{mol} \]

### Step 3: Calculate Moles of H3PO4 Produced
6. **Second Reaction**:
\[ P4O10 + 6H2O \rightarrow 4H3PO4 \]
- From the stoichiometry, 1 mole of P4O10 produces 4 moles of H3PO4.

7. **Calculate Moles of H3PO4**:
\[ \text{Moles of } H3PO4 = 4 \times \text{Moles of } P4O10 = 4 \times 0.425 = 1.7 \, \text{mol} \]

8. **Considering the Yield**:
- Given yield of H3PO4 = 90%
\[ \text{Moles obtained of } H3PO4 = \text{Moles calculated} \times \frac{\text{yield}}{100} = 1.7 \times \frac{90}{100} = 1.53 \, \text{mol} \]

### Step 4: Calculate Mass of H3PO4 Produced
9. **Molar Mass of H3PO4**:
- Molar mass of H3PO4 = 3(1) + 31 + 4(16) = 98 g/mol

10. **Calculate Mass of H3PO4**:
\[ \text{Mass of } H3PO4 = \text{Moles of } H3PO4 \times \text{Molar mass of } H3PO4 = 1.53 \, \text{mol} \times 98 \, \text{g/mol} = 149.94 \, \text{g} \]

### Final Answer
The mass of H3PO4 produced is approximately **149.94 g**.

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