What weight of HNO $_{3}$ is needed to convert $62$ gm P $_{4}$ to H $_{3}$ PO $_{4}$ in the reaction?
P $_{4}$ +HNO $_{3}$ $\to$ H $_{3}$ PO $_{4}$ +NO $_{2}$ +H $_{2}$ O

63

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630

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315

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126

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Solution

The correct option is C $630$ gm
The balanced chemical reaction is given as:
$P_{4} + 20 H N O_{3} \to 4 H_{3} P O_{4} + 20 N O_{2} + 4 H_{2} O$
Molecular weight of $P_{4} = 124 a m u$
Molecular weight of $H N O_{3} = 63 a m u$
From the balanced chemical equation, 1 mole of $P_{4}$ reacts with 20 moles of $H N O_{4}$
$∴$ 124 gm $P_{4}$ reacts with $20 \times 63 g m$ $H N O_{4}$
$\Rightarrow$ 62 gm $P_{4}$ will react with $\frac{63 \times 20 \times 62}{124} = 630 g m$ $H N O_{4}$

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Similar questions

N_{2} + 3 H_{2} \to 2 N H_{3}

4 N H_{3} + 5 O_{2} \to 4 N O + 6 H_{2} O

N_{2}

H_{2}

O_{2}

H_{3} P O_{4}

H_{3} P O_{4} + O H^{-} \to H_{2} P O_{4}^{-} + H_{2} O

H_{3} P O_{4} + 2 O H^{-} \to H P O_{4}^{2 -} + 2 H_{2} O

H_{3} P O_{4} + 3 O H^{-} \to P O_{4}^{3 -} + 3 H_{2} O

H_{3} P O_{2}, H_{3} P O_{4}, H_{3} P O_{3}

H_{4} P_{2} O_{6}

H_{2} O + H_{3} {P O}_{4} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{-}; p K_{1} = 2.15

H_{2} O + H_{2} {P O}_{4}^{-} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{2 -}; p K_{2} = 7.20

0.01 M

N a H_{2} {P O}_{4}

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