Q19 / 180 · KCET 2018 · Chemistry

\( 1.0 \mathrm{~g} \) 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?

  • A
    \( \mathrm{Mg}, 5.8 \mathrm{~g} \)
  • Correct answer
    B
    Mg, \( 0.58 \mathrm{~g} \)
  • C
    \(\mathrm{O}_{2}, 0.24 \mathrm{~g} \)
  • D
    \( \mathrm{O}_{2}, 2.4 \mathrm{~g} \)

Solution

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 $=1-0.42=0.58 \mathrm{~g}$

Hence, the correct answer is option: 'B'.

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