What weight of $HN{O_3}$ is needed to convert 62 g of ${P_4}$ into ${H_3}P{O_4}$ in the reaction?
${P_4} + HN{O_3} \to {H_3}P{O_4} + N{O_2} + {H_2}O$
A.63 g
B.630 g
C.315 g
D.126 g

Hint:Firstly, while solving, balancing of chemical reaction is important. ${P_4}$ is tetra phosphorous. It is present in three forms, they are white, red and black. We shall first balance the given equation and use the stoichiometric coefficients to calculate the weight required.

Complete step by step answer:
In the given reaction, ${P_4}$ and $HN{O_3}$ are reactants. Tetra phosphorus and nitric acid react to form phosphoric acid that is ${H_3}P{O_4}$ and nitrogen dioxide that is $N{O_2}$. ${H_2}O$ that is water is also formed. This is an oxidation reduction reaction, also known as a redox reaction. ${P_4}$ acts as a reducing agent and $HN{O_3}$ acts as an oxidizing agent in this reaction.
${P_4} + HN{O_3} \to {H_3}P{O_4} + N{O_2} + {H_2}O$
This reaction is not balanced. In the reactant side, there are four phosphorus atoms. Hence ${H_3}P{O_4}$ should be multiplied by four as there is only one phosphorus atom on the product side. And to balance the equation, $HN{O_3}$ and $N{O_2}$ should be multiplied by twenty. Also, ${H_2}O$ should be multiplied by four. After this, in the reactant side, there are four phosphorus atoms, twenty hydrogen atoms, twenty nitrogen atoms and sixty oxygen atoms which are the same as the product side.
${P_4} + 20HN{O_3} \to 4{H_3}P{O_4} + 20N{O_2} + 4{H_2}O$
One mole of ${P_4}$ reacts with twenty moles of $HN{O_3}$.
Hence 124 g of ${P_4}$ reacts with $20 \times 63$ g of $HN{O_3}$ as molar mass of $HN{O_3}$ is 63 gram per mole.
Hence 62 g of ${P_4}$ will react with x g of $HN{O_3}$.
Therefore,
$x = \dfrac{{20 \times 63 \times 62}}{{124}} = 630$
The weight of $HN{O_3}$ needed is 630 g.

Therefore, the correct option is B.

Note:
Phosphorus is an essential element which takes part in a broad variety of biochemical reactions. Aqua fortis and the spirit of niter are other names of nitric acid.