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