Assume carbon burns according to following equation :
$2 C (s) + O_{2} (g) \to 2 CO (s)$
when $12 g$ carbon is burnt in $48 g$ of oxygen, the volume of carbon monoxide produced is $______\times 10^{- 1} L$ at STP [nearest integer]

Question

[Given : Assume $CO$ as ideal gas, Mass of $C$ is $12 g {mol}^{- 1}$ , mass of $O$ is $16 g {mol}^{- 1}$ and molar volume of an idal gas at STP is $22.7 L {mol}^{- 1}$ ]

TestHub · Accepted Answer

Combustion of carbon under less supply of air,
$2 C + O_{2} \to 2 CO$ ....(X)

$\frac{12}{12}$ $\frac{48}{32}$ (Given moles of reactants)
Thus, carbon is the limiting reagent as for $1.5 mol$ of Oxygen, 3 moles of carbon is required.

Thus, from equation(X) we get,

Mole of $CO = 1$ mole

From arrhenius law,
${(V_{CO})}_{atSTP} = 22.7 Lit$

$= 227 \times 10^{- 1} Lit$

Chemistry - Mole Concept Question with Solution | TestHub

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