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The volume of oxygen at STP required to burn $2.4 \mathrm{~g}$ of carbon completely is
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Verified Answer
The correct answer is:
$4.48 \mathrm{~L}$
$C(s)+O_{2}(g) \longrightarrow C O_{2}(g)$
moles $=1$ mole 1 mole 1 mole
weight $=12 g m \quad 32 g m \quad 44 g m$
$12 g m$ of $C$ require $\rightarrow 1$ mole of $O_{2}$
volume of $2.4 / 12$ mole $\mathrm{O}_{2}$ at $\mathrm{STP}=\frac{22.4 \times 2.4}{12}$ litre
$4.48$ litre
moles $=1$ mole 1 mole 1 mole
weight $=12 g m \quad 32 g m \quad 44 g m$
$12 g m$ of $C$ require $\rightarrow 1$ mole of $O_{2}$
volume of $2.4 / 12$ mole $\mathrm{O}_{2}$ at $\mathrm{STP}=\frac{22.4 \times 2.4}{12}$ litre
$4.48$ litre
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