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Observe the following reaction
$\mathrm{xS}_8(\mathrm{~s})+\mathrm{yOH}^{-}(\mathrm{aq}) \rightarrow \mathrm{zS}^{2-}(\mathrm{aq})+2 \mathrm{~S}_2 \mathrm{O}_3^{2-}(\mathrm{aq})$
$+6 \mathrm{H}_2 \mathrm{O}(1) \mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are respectively.
Options:
$\mathrm{xS}_8(\mathrm{~s})+\mathrm{yOH}^{-}(\mathrm{aq}) \rightarrow \mathrm{zS}^{2-}(\mathrm{aq})+2 \mathrm{~S}_2 \mathrm{O}_3^{2-}(\mathrm{aq})$
$+6 \mathrm{H}_2 \mathrm{O}(1) \mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are respectively.
Solution:
2443 Upvotes
Verified Answer
The correct answer is:
$1,12,4$
The balanced equation is:-
$\begin{aligned}
& \mathrm{S}_8(\mathrm{~s})+12 \mathrm{OH}^{-} \text {(aq.) } \rightarrow \\
& 4 \mathrm{~S}^{2-}(\mathrm{aq})+2 \mathrm{~S}_2 \mathrm{O}_3{ }^{2-}(\mathrm{aq})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \\
&
\end{aligned}$
Thus, $\mathrm{x}=1, \mathrm{y}=12$ and $\mathrm{z}=4$
$\begin{aligned}
& \mathrm{S}_8(\mathrm{~s})+12 \mathrm{OH}^{-} \text {(aq.) } \rightarrow \\
& 4 \mathrm{~S}^{2-}(\mathrm{aq})+2 \mathrm{~S}_2 \mathrm{O}_3{ }^{2-}(\mathrm{aq})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \\
&
\end{aligned}$
Thus, $\mathrm{x}=1, \mathrm{y}=12$ and $\mathrm{z}=4$
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