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\( 1.0 \mathrm{~g} \) of \( \mathrm{Mg} \) is burnt with \( 0.28 \mathrm{~g} \) of \( \mathrm{O}_{2} \) in a closed vessel. Which reactant is left in excess and
how much?
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how much?
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Verified Answer
The correct answer is:
Mg, \( 0.58 \mathrm{~g} \)
For the reaction
$2 \mathrm{Mg}+\mathrm{O}_{2} \rightarrow 2 \mathrm{M} \mathrm{gO}$
$48 \mathrm{~g}$ of $\mathrm{Mg}$ requires $32 \mathrm{~g}$ of $\mathrm{O}_{2}$
Therefore, $1 \mathrm{~g}$ of $\mathrm{Mg}$ require $\frac{32}{48} \mathrm{~g}$ of $\mathrm{O}_{2}=0.66 \mathrm{~g}$
But available $\mathrm{O}_{2}$ is $0.28 \mathrm{~g}$
So, $\mathrm{O}_{2}$ is the limiting reagent and $\mathrm{Mg}$ is the excess reagent.
$32 \mathrm{~g}$ of $\mathrm{O}_{2}$ needs $48 \mathrm{~g}$ of $\mathrm{Mg}$
$1 \mathrm{~g}$ of $\mathrm{O}_{2}$ needs $\frac{48}{32} \mathrm{~g}$ of $\mathrm{Mg}$
$0.28$ of $\mathrm{O}_{2}$ needs $\frac{48}{32} \times 0.28=0.42 \mathrm{~g}$
Mg available $=1 \mathrm{~g}$
Mg left $=1-0.42=0.58 \mathrm{~g}$
$2 \mathrm{Mg}+\mathrm{O}_{2} \rightarrow 2 \mathrm{M} \mathrm{gO}$
$48 \mathrm{~g}$ of $\mathrm{Mg}$ requires $32 \mathrm{~g}$ of $\mathrm{O}_{2}$
Therefore, $1 \mathrm{~g}$ of $\mathrm{Mg}$ require $\frac{32}{48} \mathrm{~g}$ of $\mathrm{O}_{2}=0.66 \mathrm{~g}$
But available $\mathrm{O}_{2}$ is $0.28 \mathrm{~g}$
So, $\mathrm{O}_{2}$ is the limiting reagent and $\mathrm{Mg}$ is the excess reagent.
$32 \mathrm{~g}$ of $\mathrm{O}_{2}$ needs $48 \mathrm{~g}$ of $\mathrm{Mg}$
$1 \mathrm{~g}$ of $\mathrm{O}_{2}$ needs $\frac{48}{32} \mathrm{~g}$ of $\mathrm{Mg}$
$0.28$ of $\mathrm{O}_{2}$ needs $\frac{48}{32} \times 0.28=0.42 \mathrm{~g}$
Mg available $=1 \mathrm{~g}$
Mg left $=1-0.42=0.58 \mathrm{~g}$
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