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Question: What weight of \({{P}_{4}}{{O}_{6}}\) and \({{P}_{4}}{{O}_{10}}\) will be produced by the combustion...

What weight of P4O6{{P}_{4}}{{O}_{6}} and P4O10{{P}_{4}}{{O}_{10}} will be produced by the combustion of 31 grams of P4{{P}_{4}} with 32 grams of oxygen, leaving no P4{{P}_{4}} and noO2{{O}_{2}}?
(A) 2.75g, 219.5g
(B) 27.5g, 35.5g
(C) 55g, 71g
(D) 17.5g, 190.5g

Explanation

Solution

Write down the balanced chemical equation of combustion of P4{{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=wM or n×M=wn=\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 P4{{P}_{4}} :-
P4+3O2→P4O6{{P}_{4}}+3{{O}_{2}}\to {{P}_{4}}{{O}_{6}}
From the above reaction we can see that 1 mole of P4{{P}_{4}} reacts with 3 moles of O2{{O}_{2}} to give 1 mole of P4O6{{P}_{4}}{{O}_{6}}.
- Calculation of no. of moles:-
Given mass of P4{{P}_{4}}=31g
Molar mass of P4{{P}_{4}}=4×31g/mol4\times 31g/mol
Therefore total no. of moles of P4{{P}_{4}}= n=31g4×31g/mol=0.25 moles n=\dfrac{31g}{4\times 31g/mol}\text{=0}\text{.25 moles }
Given mass of O2{{O}_{2}}=32g
Molar mass of O2{{O}_{2}}=32g/mol32g/mol
Therefore total no. of moles of O2{{O}_{2}}= n=32g32g/mol=1mole n=\dfrac{32g}{32g/mol}\text{=1}\text{mole }
-Finding the limiting reagent of this reaction:-
Since 1 mole of P4{{P}_{4}} reacts with 3 moles of O2{{O}_{2}} then 0.25 moles of P4{{P}_{4}} will react with= 0.25 moles of P4×3 moles of O21 mole of P4=0.75moles of O20.25\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 P4{{P}_{4}} is the limiting reagent and only 0.75 moles of O2{{O}_{2}} will be consumed.
- Calculating the no. of moles of P4O6{{P}_{4}}{{O}_{6}} produced:-
Since 1 mole of P4{{P}_{4}} produces 1 mole of P4O6{{P}_{4}}{{O}_{6}} then 0.25 moles of P4{{P}_{4}} produces= 0.25 moles of P4×1 mole of P4O6 1 mole of P4=0.25moles of P4O60.25\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 P4O6{{P}_{4}}{{O}_{6}}:-
P4O6+2O2→P4O10{{P}_{4}}{{O}_{6}}+2{{O}_{2}}\to {{P}_{4}}{{O}_{10}}
From the reaction we can see that 1 mole of P4O6{{P}_{4}}{{O}_{6}} reacts with 2 moles of O2{{O}_{2}} to give 1 mole ofP4O10{{P}_{4}}{{O}_{10}}.
-Finding the limiting reagent of this reaction:-
Since 1 mole of P4O6{{P}_{4}}{{O}_{6}} reacts with 2 moles of O2{{O}_{2}} then 0.25 moles of P4O6{{P}_{4}}{{O}_{6}} will react with=
0.25 moles of P4O6×2 moles of O21 mole of P4O6=0.50moles of O20.25\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 O2{{O}_{2}} were already consumed and only 0.25 moles were left. But for reaction with 0.25 moles of P4O6{{P}_{4}}{{O}_{6}}, we require 0.50 moles of O2{{O}_{2}} which is not available. Therefore O2{{O}_{2}} will be the limiting reagent this time and we will calculate the no. of moles of P4O10{{P}_{4}}{{O}_{10}} produced with the help of it.
-Calculating the no. of moles of P4O6{{P}_{4}}{{O}_{6}} produced:-
Since 2 mole ofO2{{O}_{2}} produces 1 mole of P4O10{{P}_{4}}{{O}_{10}} then 0.25 moles of O2{{O}_{2}} produces= 0.25 moles of O2×1 mole of P4O10 2 mole of O2=0.125moles of P4O100.25\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 P4O6{{P}_{4}}{{O}_{6}} left:-
Consumed moles of P4O6{{P}_{4}}{{O}_{6}}=0.25 moles of O2×1 mole of P4O6 2 mole of O2=0.125moles of P4O60.25\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 P4O6{{P}_{4}}{{O}_{6}}left= 0.25-0.125=0.125 moles
-Calculation of mass produced of P4O6{{P}_{4}}{{O}_{6}} and P4O10{{P}_{4}}{{O}_{10}}:-
Molecular mass of P4O6{{P}_{4}}{{O}_{6}}= 31×4+16×6=220g/mol31\times 4+16\times 6=220g/mol
Molecular mass of P4O10{{P}_{4}}{{O}_{10}}=31×4+16×10=284g/mol31\times 4+16\times 10=284g/mol
Weight (mass) of P4O6{{P}_{4}}{{O}_{6}} produced w=n×M=0.125mol×220g/mol=27.5gw=n\times M=0.125mol\times 220g/mol=27.5g
Weight (mass) of P4O10{{P}_{4}}{{O}_{10}} produced w=n×M=0.125mol×284g/mol=35.5gw=n\times M=0.125mol\times 284g/mol=35.5g
Also all the mass of P4{{P}_{4}} and O2{{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.