Answer
396.9k+ views
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.
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.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)