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Phosphoric acid $\left( {{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}} \right)$ is widely used to make fertilizers and can be prepared in a two-step process.
Step I: ${{\text{P}}_{\text{4}}}\, + \,{\text{5}}\,{{\text{O}}_{\text{2}}}\, \to \,{{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$
Step II: ${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}\, + \,{\text{6}}\,{{\text{H}}_{\text{2}}}{\text{O}}\, \to \,4\,{{\text{H}}_3}{\text{P}}{{\text{O}}_{\text{4}}}$
We allow $310$ g of phosphorus to react with excess oxygen, which forms tetraphosphorus oxide $\left( {{{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}} \right)$in $50$% yield. In the step II, $25$% yield of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}$is obtained.
(atomic weight in amu of P = $31$, H = $1$, O = $16$)
Assuming $100$% yield, Calculate the number of moles of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}$ ?

Last updated date: 14th Jun 2024
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Hint:To determine the number of moles and grams of any reactant or product we need a balanced chemical equation. After writing the balanced equation, by comparing the number of moles of the reactant and product, we will determine the mole of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}$. We can determine the number of mole of tetraphosphorus oxide $\left( {{{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}} \right)$by using the mole formula.

Formula used:${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$

Complete step-by-step answer:We will determine the mole of phosphorus as follows:

It is given to us that $310$ g of phosphorus react. The molar mass of ${{\text{P}}_{\text{4}}}$ is $124$amu.
So, the mole of ${{\text{P}}_{\text{4}}}$is,
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
On substituting $310$ g for mass and $124$amu for molar mass of ${{\text{P}}_{\text{4}}}$,
${\text{Mole}}\,{\text{ = }}\,\dfrac{{310}}{{{\text{124}}}}$
${\text{Mole}}\,{\text{ = }}\,2.5$
Now, we will compare the moles of phosphorus with moles of tetraphosphorus dioxide as follows:
According to the step I Step I: ${{\text{P}}_{\text{4}}}\, + \,{\text{5}}\,{{\text{O}}_{\text{2}}}\, \to \,{{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$ , one mole of ${{\text{P}}_{\text{4}}}$is giving one mole of ${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$ so, $2.5$moles ${{\text{P}}_{\text{4}}}$will give $2.5$moles ${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$ but the yield of product is $50$% it means one mole of ${{\text{P}}_{\text{4}}}$ gives $0.5$mole of ${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$ so, $2.5$moles ${{\text{P}}_{\text{4}}}$will give,
$1$ Mole of ${{\text{P}}_{\text{4}}}$= $0.5$mole of ${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$
$2.5$moles ${{\text{P}}_{\text{4}}}$= $1.25$mole of ${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$
So, we have $1.25$mole of ${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$for the reaction of step II.
According to step II, ${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}\, + \,{\text{6}}\,{{\text{H}}_{\text{2}}}{\text{O}}\, \to \,4\,{{\text{H}}_3}{\text{P}}{{\text{O}}_{\text{4}}}$, one mole of ${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$ is giving $4$moles of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}$and it is given that we have to assume$100$% yield, means completely reactant${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$ is converting into product. So, $1.25$mole of ${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$will give,
$1$ mole of ${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$ = $4$moles of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}$
$1.25$ mole of ${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$ = $5$moles of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}$
So, $1.25$ mole of ${{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}$ will give $5$moles of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}$
Therefore, the number of moles of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_4}$is $5$moles.

Note:Stoichiometry measurements are used to determine the amount of reactant or product from the given amounts. Stoichiometry measurements give quantitative relations among the amounts of various species of a reaction. To determine the stoichiometry relations a balanced equation is necessary. Percent yield is determined as the actual amount of product divided by the theoretical yield of product and multiplied with hundred. The molar mass of the compound is determined by adding the atomic mass of constituting atoms.