Question #6ff98

Jul 5, 2016

$1.87 L$

Explanation:

It may be easily solved by using mole concept as follows:
I assumed that the reactants are completely consumed to form both
$C O$ & $C {O}_{2}$
and they are formed as

$C O \to x \text{ moles}$

$C {O}_{2} \to y \text{ moles}$

We have been given reatants as

$C \to 2 g \to \text{2g"/"12g/mole"=1/6"moles}$

${O}_{2} \to 4 g \to \text{4g"/"32g/mol"=1/8"moles}$

Now 1 mole CO contains 1 mole C and $\frac{1}{2}$ mole ${O}_{2}$

So x mole CO contains x mole C and $\frac{x}{2}$ mole ${O}_{2}$

Similarly 1 mole $C {O}_{2}$ contains 1 mole C and 1 mole ${O}_{2}$

So y mole $C {O}_{2}$ contains y mole C and y mole ${O}_{2}$

We can cobmine these information to form following two equations

Considering Total Carbon

$x + y = \frac{1}{6.} \ldots . . \left(1\right)$

Considering Total Oxygen

$\frac{x}{2} + y = \frac{1}{8.} \ldots . \left(2\right)$

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

$x - \frac{x}{2} = \frac{1}{6} - \frac{1}{8} = \frac{4 - 3}{24}$

$\therefore \frac{x}{2} = \frac{1}{24} \implies x = \frac{1}{12} m o l$

Plugging the value of x in (1)

$\frac{1}{12} + y = \frac{1}{6}$

$\implies y = \frac{1}{6} - \frac{1}{12} = \frac{2 - 1}{12}$

$\therefore y = \frac{1}{12} m o l$

When the mixture of gas produced is passed through alkali solution $C {O}_{2}$ will be absorbed and the residual gas will be $\frac{1}{12} m o l$ CO which will occupy $\frac{1}{12} \times 22.4 L \approx 1.87 L \text{ at STP}$