Question #100b4

1 Answer
Jun 8, 2017

Molar masses of

C_2H_2to2xx12+2xx1=26g"/"mol

C_2H_4to2xx12+4xx1=28g"/"mol

CH_4to1xx12+4xx1=16g"/"mol

The mole ratio in the mixture

C_2H_2:C_2H_4:CH_4=2:1:2

The mass ratio in the mixture

C_2H_2:C_2H_4:CH_4=(2xx26):(1xx28):(2xx16)=13:7:8

The masses of the components in the gas mixture of 1g

C_2H_2to13/28g-> 13/28xx1/26mol=1/56mol

C_2H_4to7/28g->7/28xx1/28mol=1/112mol

CH_4to8/28g->8/28xx1/16mol=1/56mol

Balanced equation of the reactions

C_2H_2(g)+5/2O_2(g)->2CO_2(g)+H_2O(l)

C_2H_4(g)+3O_2(g)->2CO_2(g)+2H_2O(l)

CH_4(g)+2O_2(g)->CO_2(g)+2H_2O(l)

So the number of moles of color(red)(O_2(g)) required for complete burning of component gases are

C_2H_2to1/56xx5/2=5/112mol

C_2H_4to1/112xx3=3/112mol

CH_4to1/56xx2=4/112mol

So total amount of color(red)(O_2(g)) required for complete burning of 1 g gas mixture is

O_2->(5+3+4)/112mol=12/112mol

As the ratio of O_2 :N_2 by volume in air is (20%):(80%)=1:4

The mole ratio of O_2 :N_2 in air will be =1:4

So in air 12/112mol O_2 will be with 12/112xx4mol N_2

Hence total mass of air required for complete burning is

12/112mol O_2+ 12/112xx4mol N_2

=(12xx32+48xx28)/112g=1728/112g-> option A