Question

# The mole fraction of $$C_2H_4$$ in the mixture is :

A
0.25
B
0.75
C
0.45
D
0.55

Solution

## The correct option is A $$0.75$$The balanced reactions are as follows:$${C_3H_8}+{5O_2}\rightarrow {3CO_2}+{4H_2O(l)}$$$${C_3H_4}+{3O_2}\rightarrow {2CO_2}+{4H_2O(l)}$$Initially,$$PV=nRT$$$$4.93\times V=(a+b)RT$$ .....(i)After combustion, pressure is due to the total moles of $$CO_2$$.$$11.8\times V=(3a+2b)RT$$ .....(ii)Divide equation (ii) by equation (i), we get$$\dfrac {11.08}{4.93}=2.25=\dfrac {3a+2b}{a+b}$$$$2.25a+2.25b=3a+2b$$$$0.25b=0.75a$$$$b=3a$$$$\chi_{C_2H_8}$$ or $$\chi_a=\dfrac {a}{a+b}=\dfrac {a}{a+3a}=\dfrac {1}{4}=0.25$$$$\chi_{C_2H_4}$$ or $$\chi_b=1-0.25=0.75$$Hence, the mole fraction of $$C_2H_4$$ in the mixture is $$0.75$$.Chemistry

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