Answer
384k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 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)
Give 10 examples for herbs , shrubs , climbers , creepers
![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)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english 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)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)