Question #6ff98

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Question #6ff98

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Answer 1 · 2016-07-05T08:05:54

$1.87 L$ $1.87L$

Explanation:

It may be easily solved by using mole concept as follows:
I assumed that the reactants are completely consumed to form both
$C O$ $CO$ & $C O_{2}$ $CO_2$
and they are formed as

$C O \to x moles$ $CO->x" moles"$

$C O_{2} \to y moles$ $CO_2->y" moles"$

We have been given reatants as

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

$O_{2} \to 4 g \to \frac{4g}{32g/mol} = \frac{1}{8} moles$ $O_2->4g->"4g"/"32g/mol"=1/8"moles"$

Now 1 mole CO contains 1 mole C and $\frac{1}{2}$ $1/2$ mole $O_{2}$ $O_2$

So x mole CO contains x mole C and $\frac{x}{2}$ $x/2$ mole $O_{2}$ $O_2$

Similarly 1 mole $C O_{2}$ $CO_2$ contains 1 mole C and 1 mole $O_{2}$ $O_2$

So y mole $C O_{2}$ $CO_2$ contains y mole C and y mole $O_{2}$ $O_2$

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 $1/12xx22.4L~~ 1.87L" at STP"$

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Question #6ff98

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