Question #6ff98

1 Answer
Jul 5, 2016

Answer:

#1.87L#

Explanation:

It may be easily solved by using mole concept as follows:
I assumed that the reactants are completely consumed to form both
#CO # & # CO_2#
and they are formed as

#CO->x" moles"#

#CO_2->y" moles"#

We have been given reatants as

#C->2g->"2g"/"12g/mole"=1/6"moles"#

#O_2->4g->"4g"/"32g/mol"=1/8"moles"#

Now 1 mole CO contains 1 mole C and #1/2# mole #O_2#

So x mole CO contains x mole C and #x/2# mole #O_2#

Similarly 1 mole #CO_2# contains 1 mole C and 1 mole #O_2#

So y mole #CO_2# contains y mole C and y mole #O_2#

We can cobmine these information to form following two equations

Considering Total Carbon

#x+y=1/6......(1)#

Considering Total Oxygen

#x/2+y=1/8.....(2)#

Now subtracting (2) from (1) we can write

#x-x/2=1/6-1/8=(4-3)/24#

#:.x/2=1/24=>x=1/12mol#

Plugging the value of x in (1)

#1/12+y=1/6#

#=>y=1/6-1/12=(2-1)/12#

#:.y=1/12mol#

When the mixture of gas produced is passed through alkali solution #CO_2# will be absorbed and the residual gas will be #1/12mol# CO which will occupy #1/12xx22.4L~~ 1.87L" at STP"#