When $12 g$ graphite is burnt in sufficient oxygen, $C O$ as well as $C O_{2}$ is formed. If the product contains $40$ % $C O$ and $60$ % $C O_{2}$ by mass and none of the reactant is left, what is the mass of oxygen gas used in combustion?

24.0 g

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21.12 g

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23.8 g

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15.6 g

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Solution

The correct option is B $21.12 g$
Given,
$12 g$ graphite is burnt to give $C O$ & $C O_{2}$
The reaction involved is
$3 C + 2 O_{2} ⟶ 2 C O + C O_{2}$
Also, given that no reactant is left over
$\Rightarrow$ The reaction is complete
From above data
Moles of graphite given= $\frac{12 g}{12 g / m o l e} = 1$ mole
From above reaction, it is clear that $3$ moles of $C$ require $2$ moles of $O_{2}$ for complete oxidization.
$\Rightarrow 1$ mole of $C$ require, moles of $O_{2} = \frac{2}{3} = 0.66$ moles
We know $1$ mole $O_{2} = 32 g$
$\Rightarrow 0.66$ mole $O_{2} = 0.66 \times 32$
$= 21.12 g$

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When 12 g graphite is burnt in sufficient oxygen, CO as well as CO2 is formed. If the product contains 40% CO and 60% CO2 by mass and none of the reactant is left, what is the mass of oxygen gas used in combustion?

When $12 g$ graphite is burnt in sufficient oxygen, $C O$ as well as $C O_{2}$ is formed. If the product contains $40$ % $C O$ and $60$ % $C O_{2}$ by mass and none of the reactant is left, what is the mass of oxygen gas used in combustion?