Question

Question #6ff98

Answer 1

`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"`