Question

In the following chemical equation, the co-efficient of O2 in accordance with the law of conservation of mass is:- C₅H1₂ + O₂ ------------> 5CO₂ + 6H₂O.

Please give full steps about how to find the answer
(plz answer according to class 9th standard CBSE )

Answer 1

The equation given is balanced with respect to C and H . we can find the coefficient by just counting and balancing the no of oxygen atoms .
But question demands that we have to use Law of conservation of mass .
According to Law of conservation of mass ,
Mass of Reactants = Mass of Products
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#Mass of Reactants :-
Mass of 1 mol of C5H12 = 72 g
Suppose coefficient of O2 is 'x' then,
Mass of 'x' mol of O2 = 32x g
( as 1 mol of O2 has 32 g mass)
Total mass of Reactants = 72+32x

#Mass of Products :
Mass of 5 moles of CO2 = 5×44 =220 g
Mass of 6 moles of H2O = 6×18 =108 g
Total mass of Products = 220+108 .
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Further ;
using Law of conservation of mass :
72+32x = 220+108
on solving we find coefficient ( x) of O2
= 8 .
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hope it helps!

Answer 2

I give general solution for this ,

CxHy +(x + y/4) O2 ----->x CO2 +y/2H2O

where x and y are integer ,
now, put x = 5
and y = 12

then,
C5H12 +(5 + 12/4 )O2 ----->5CO2 +(12/2)H2O

C5H12 +8O2 --->5CO2 + 6H2O
in this way you easily find out balance chemical equation of any hydrocarbon racted with oxygen .