10 moles $\mathrm{SO}_{2}$ and 15 moles $\mathrm{O}_{2}$ were allowed to react over a suitable catalyst. 8…

10 moles $\mathrm{SO}_{2}$ and 15 moles $\mathrm{O}_{2}$ were allowed to react over a suitable catalyst. 8 moles of $\mathrm{SO}_{3}$ were formed. The remaining moles of $\mathrm{SO}_{2}$ and $\mathrm{O}_{2}$ respectively are -
  1. 2 moles, 11 moles
  2. 2 moles, 8 moles
  3. 4 moles, 5 moles
  4. 8 moles, 2 moles

Solution

$2 \mathrm{SO}_{2} \quad+\quad \mathrm{O}_{2} \longrightarrow 2 \mathrm{SO}_{3}$
$\begin{array}{lll}10 & 15 & 0 \\ 10-2 x & 15-x & 2 x\end{array}$
$\because 2 x=8 \quad x=4$
Hence, remaining, $\mathrm{SO}_{2}=10-8=2$ moles,
$\mathrm{O}_{2}=15-4=11$ moles

Asked in: JEE-TOPICTESTS-CHEMISTRY

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