A balanced equation for the conversion of phosphorus to phosphoric acid is given by :

P + {5 H N O}_{3} \to H_{2} O + H_{3} {P O}_{2} + {5 N O}_{2}

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P + {5 H N O}_{3} \to H_{2} O + H_{2} {P O}_{5} + {5 N O}_{2}

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P + {3 H N O}_{3} \to H_{2} O + H_{3} {P O}_{4} + {3 N O}_{2}

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P + {5 H N O}_{3} \to H_{2} O + H_{3} {P O}_{4} + {5 N O}_{2}

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Solution

The correct option is C $P + {5 H N O}_{3} \to H_{2} O + H_{3} {P O}_{4} + {5 N O}_{2}$
Oxidation of red phosphorous $(P)$ by nitric acid $(H N O_{3})$ yields phosphoric acid $(H_{3} P O_{4})$ .

$P + {5 H N O}_{3} \to H_{2} O + H_{3} {P O}_{4} + {5 N O}_{2}$

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Similar questions

P + 5 H N O_{3} ⟶ H_{3} P O_{4} + H_{2} O + 5 N O_{2}

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

S T P

[H = 1; N = 14; O = 16; P = 31]

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}

P + 5 H N O_{3} ⟶ H_{3} P O_{4} + H_{2} O + 5 N O_{2}

N O_{2}

H

N

P

O

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

H N O_{3}

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