What weight of `HNO_(3)` is needed to convert 62gm of `P_(4)` in `H_(3) PO_(4) ` in the region ?
`P_(4) + HNO_(3) rarr H_(3) PO_(4) + NO_(2) + H_(2) O`

Options

A. 63gm
B. 630gm
C. 315 gm
D. 126 gm

Explanation

To solve the problem of how much weight of HNO₃ is needed to convert 62 g of P₄ into H₃PO₄, we will follow these steps:

### Step 1: Write the balanced chemical equation
The unbalanced reaction is:
\[ P_4 + HNO_3 \rightarrow H_3PO_4 + NO_2 + H_2O \]

The balanced equation is:
\[ P_4 + 20 HNO_3 \rightarrow 4 H_3PO_4 + 20 NO_2 + 4 H_2O \]

### Step 2: Determine the molar masses
- Molar mass of P₄:
\[ \text{Molar mass of P} = 31 \, \text{g/mol} \]
\[ \text{Molar mass of P}_4 = 31 \times 4 = 124 \, \text{g/mol} \]

- Molar mass of HNO₃:
\[ \text{Molar mass of HNO}_3 = 1 + 14 + (16 \times 3) = 63 \, \text{g/mol} \]

### Step 3: Calculate the number of moles of P₄
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
For 62 g of P₄:
\[ \text{Number of moles of P}_4 = \frac{62 \, \text{g}}{124 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 4: Determine the moles of HNO₃ required
From the balanced equation, we see that 1 mole of P₄ requires 20 moles of HNO₃. Therefore, for 0.5 moles of P₄:
\[ \text{Moles of HNO}_3 = 0.5 \, \text{mol} \times 20 = 10 \, \text{mol} \]

### Step 5: Calculate the mass of HNO₃ required
Using the number of moles and the molar mass:
\[ \text{Mass of HNO}_3 = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Mass of HNO}_3 = 10 \, \text{mol} \times 63 \, \text{g/mol} = 630 \, \text{g} \]

### Final Answer
The weight of HNO₃ needed to convert 62 g of P₄ into H₃PO₄ is **630 g**.

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