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Acidified $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ solution turms green when $\mathrm{Na}_2 \mathrm{SO}_3$ is added to it. This is due to the formation of
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Verified Answer
The correct answer is:
$\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$
$$
\begin{array}{r}
\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+3 \mathrm{Na}_2 \mathrm{SO}_3+4 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow 3 \mathrm{Na}_2 \mathrm{SO}_4 \\
+\mathrm{K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+4 \mathrm{H}_2 \mathrm{O}
\end{array}
$$
\begin{array}{r}
\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+3 \mathrm{Na}_2 \mathrm{SO}_3+4 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow 3 \mathrm{Na}_2 \mathrm{SO}_4 \\
+\mathrm{K}_2 \mathrm{SO}_4+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+4 \mathrm{H}_2 \mathrm{O}
\end{array}
$$
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