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Question
Concentrated nitric acid oxidizes phosphorus to phosphoric acid according to the following equation:
\[\ce{P + 5HNO3 -> H3PO4 + 5NO2 + H2O}\]
If 6.2 g of phosphorus was used in the reaction, calculate the number of moles of phosphorus taken and mass of phosphoric acid formed.
Numerical
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Solution
\[\ce{\underset{[Steam 1 mole]}{P + 5HNO3} -> H3PO4 + H2O + 5NO2}\]
1 mole of phosphorus atom = 31 g of phosphorus
31 g of P = 1 mole of P
∴ 6.2 g of P = `(6.2 xx 1)/31`
= 0.2 mole of P
Also, 31 g of phosphorus or 1 mole of phosphorus produces H3PO4
= (3 × 1) + 31 + (4 × 16)
= 98 g of H3PO4
∴ 0.2 mole of P produces = 98 × 0.2
= 19.6 g of H3PO4
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