Question

Calculate the mass of phosphoric acid required to obtain $53.4gm$ of pyrophosphoric acid.

Answer 1

To solve this question, we need to know that pyrophosphoric acid can be obtained from phosphoric acid and to obtain one mole of pyrophosphoric acid, two moles of phosphoric acid are required.

Complete step by step answer:
In this question, pyrophosphoric acid has the chemical formula ${H_3}{P_2}{O_7}$ and phosphoric acid has the chemical formula ${H_3}P{O_4}$. The pyrophosphoric acid can be obtained from phosphoric acid by the following given reaction:
$2{H_3}P{O_4} \to {H_3}{P_2}{O_7} + {H_2}O$
From the following equation we can say that to obtain one mole of pyrophosphoric acid, two moles of phosphoric acid are required.
Now, we will find a mass of phosphoric acid in one mole that is molar mass. As in phosphoric acid, there are three hydrogen and mass of one hydrogen is $1gm$, there is one phosphorus whose atomic mass is $31gm$ and four oxygen atoms and mass of one oxygen atom is $16gm$.
Now, Molecular mass of one mole phosphoric acid $= 3 \times 1 + 31 \times 1 + 16 \times 4$
$= 98gm$
Mass of $2moles$ of phosphoric acid $= 2 \times 98$
$= 196gm$
Now,
Mass of $1mole$of pyrophosphoric acid $= 3 \times 1 + 2 \times 31 + 7 \times 16$
$= 177.9gm$
Now, as $1mole$ of pyrophosphoric acid is obtained from $2moles$ of phosphoric acid
It means, $196gm$ of phosphoric acid will be required for pyrophosphoric acid.
Now, $1gm$ of pyrophosphoric acid will require $\dfrac{{196}}{{177.9}}gm$ of phosphoric acid
Now, $1gm$ of pyrophosphoric acid will require $1.102gm$ of phosphoric acid
As given in question for $53.4gm$ of pyrophosphoric acid,
Now, $53.4gm$ of pyrophosphoric acid will require $1.102 \times 53.4gm$ of phosphoric acid
Hence, $53.4gm$ of pyrophosphoric acid will require $58.8gm$ of phosphoric acid.

Note:
In such questions, we need to remember that two moles of phosphoric acid are required to produce one mole of pyrophosphoric acid in the product and also the chemical formula of pyrophosphoric acid that is ${H_3}{P_2}{O_7}$.