Question

What weight of ${{P}_{4}}{{O}_{6}}$ and ${{P}_{4}}{{O}_{10}}$ will be produced by the combustion of 31 grams of ${{P}_{4}}$ with 32 grams of oxygen, leaving no ${{P}_{4}}$ and no${{O}_{2}}$?
(A) 2.75g, 219.5g
(B) 27.5g, 35.5g
(C) 55g, 71g
(D) 17.5g, 190.5g

Answer 1

Hint: Write down the balanced chemical equation of combustion of ${{P}_{4}}$ and find out the limiting reagent. Then, use this limiting reagent to calculate the moles of product generated, followed by calculation of their masses.

Formula used: We will use the following formula at various steps in this solution:-
$n=\dfrac{w}{M}\text{ or }n\times M=w$
where,
n= no. of moles
w= given mass or mass produced
M= molecular mass

Complete step-by-step answer:
Let us first write down the balanced chemical equation and solve it in steps as follows:-
- Combustion of ${{P}_{4}}$ :-
${{P}_{4}}+3{{O}_{2}}\to {{P}_{4}}{{O}_{6}}$
From the above reaction we can see that 1 mole of ${{P}_{4}}$ reacts with 3 moles of ${{O}_{2}}$ to give 1 mole of ${{P}_{4}}{{O}_{6}}$.
- Calculation of no. of moles:-
Given mass of ${{P}_{4}}$=31g
Molar mass of ${{P}_{4}}$=$4\times 31g/mol$
Therefore total no. of moles of ${{P}_{4}}$= $n=\dfrac{31g}{4\times 31g/mol}\text{=0}\text{.25 moles }$
Given mass of ${{O}_{2}}$=32g
Molar mass of ${{O}_{2}}$=$32g/mol$
Therefore total no. of moles of ${{O}_{2}}$= $n=\dfrac{32g}{32g/mol}\text{=1}\text{mole }$
-Finding the limiting reagent of this reaction:-
Since 1 mole of ${{P}_{4}}$ reacts with 3 moles of ${{O}_{2}}$ then 0.25 moles of ${{P}_{4}}$ will react with= $0.25\text{ moles of }{{\text{P}}_{4}}\times \dfrac{\text{3 moles of }{{\text{O}}_{2}}}{\text{1 mole of }{{\text{P}}_{4}}}=0.75\text{moles of }{{\text{O}}_{2}}$
Hence, in this reaction ${{P}_{4}}$ is the limiting reagent and only 0.75 moles of ${{O}_{2}}$ will be consumed.
- Calculating the no. of moles of ${{P}_{4}}{{O}_{6}}$ produced:-
Since 1 mole of ${{P}_{4}}$ produces 1 mole of ${{P}_{4}}{{O}_{6}}$ then 0.25 moles of ${{P}_{4}}$ produces= $0.25\text{ moles of }{{\text{P}}_{4}}\times \dfrac{\text{1 mole of }{{\text{P}}_{4}}{{\text{O}}_{6}}\text{ }}{\text{1 mole of }{{\text{P}}_{4}}}=0.25\text{moles of }{{\text{P}}_{4}}{{\text{O}}_{6}}$
- Combustion of ${{P}_{4}}{{O}_{6}}$:-
${{P}_{4}}{{O}_{6}}+2{{O}_{2}}\to {{P}_{4}}{{O}_{10}}$
From the reaction we can see that 1 mole of ${{P}_{4}}{{O}_{6}}$ reacts with 2 moles of ${{O}_{2}}$ to give 1 mole of${{P}_{4}}{{O}_{10}}$.
-Finding the limiting reagent of this reaction:-
Since 1 mole of ${{P}_{4}}{{O}_{6}}$ reacts with 2 moles of ${{O}_{2}}$ then 0.25 moles of ${{P}_{4}}{{O}_{6}}$ will react with=
$0.25\text{ moles of }{{\text{P}}_{4}}{{\text{O}}_{6}}\times \dfrac{\text{2 moles of }{{\text{O}}_{2}}}{\text{1 mole of }{{\text{P}}_{4}}{{\text{O}}_{6}}}=0.50\text{moles of }{{\text{O}}_{2}}$
Earlier we saw that 0.75 moles of ${{O}_{2}}$ were already consumed and only 0.25 moles were left. But for reaction with 0.25 moles of ${{P}_{4}}{{O}_{6}}$, we require 0.50 moles of ${{O}_{2}}$ which is not available. Therefore ${{O}_{2}}$ will be the limiting reagent this time and we will calculate the no. of moles of ${{P}_{4}}{{O}_{10}}$ produced with the help of it.
-Calculating the no. of moles of ${{P}_{4}}{{O}_{6}}$ produced:-
Since 2 mole of${{O}_{2}}$ produces 1 mole of ${{P}_{4}}{{O}_{10}}$ then 0.25 moles of ${{O}_{2}}$ produces= $0.25\text{ moles of }{{\text{O}}_{2}}\times \dfrac{\text{1 mole of }{{\text{P}}_{4}}{{\text{O}}_{10}}\text{ }}{\text{2 mole of }{{\text{O}}_{2}}}=0.125\text{moles of }{{\text{P}}_{4}}{{\text{O}}_{10}}$
-Calculating the no. of moles of ${{P}_{4}}{{O}_{6}}$ left:-
Consumed moles of ${{P}_{4}}{{O}_{6}}$=$0.25\text{ moles of }{{\text{O}}_{2}}\times \dfrac{\text{1 mole of }{{\text{P}}_{4}}{{\text{O}}_{6}}\text{ }}{\text{2 mole of }{{\text{O}}_{2}}}=0.125\text{moles of }{{\text{P}}_{4}}{{\text{O}}_{6}}$
No. of moles of ${{P}_{4}}{{O}_{6}}$left= 0.25-0.125=0.125 moles
-Calculation of mass produced of ${{P}_{4}}{{O}_{6}}$ and ${{P}_{4}}{{O}_{10}}$:-
Molecular mass of ${{P}_{4}}{{O}_{6}}$= $31\times 4+16\times 6=220g/mol$
Molecular mass of ${{P}_{4}}{{O}_{10}}$=$31\times 4+16\times 10=284g/mol$
Weight (mass) of ${{P}_{4}}{{O}_{6}}$ produced $w=n\times M=0.125mol\times 220g/mol=27.5g$
Weight (mass) of ${{P}_{4}}{{O}_{10}}$ produced $w=n\times M=0.125mol\times 284g/mol=35.5g$
Also all the mass of ${{P}_{4}}$ and ${{O}_{2}}$ is consumed as well.

Therefore the correct option is: (B) 27.5g, 35.5g

Note: -For such questions, always write a balanced chemical equation and try to find a limiting reagent at the beginning otherwise the mass of products calculated will be wrong.
-Kindly convert all values into the same units and preferably do calculations along with the units to get the desired result.