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In a balanced equation $\mathrm{H}_2 \mathrm{SO}_4+x \mathrm{HI} \rightarrow \mathrm{H}_2 \mathrm{~S}+y \mathrm{I}_2+z \mathrm{H}_2 \mathrm{O}$, the values of $x, y, z$ are
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Verified Answer
The correct answer is:
$x=8, y=4, z=4$
To balance the number of iodide atoms, HI is multiplied by 2. Iodine is oxidised with oxidation state changing from -1 to 0.
Net increase in electron is '2' Sulphur is reduced, oxidation state changing from +6 to \(-2 \Rightarrow\) electrons
To balance 8 electrons, multiply \(H I\) and \(I_2\) by 4
Hence,
\(\begin{array}{ll}
\mathrm{H}_2 \mathrm{SO}_4+x \mathrm{HI} \longrightarrow \mathrm{H}_2 \mathrm{~S}+y \mathrm{I}_2+2 \mathrm{H}_2 \mathrm{O} \\
\Rightarrow \mathrm{H}_2 \mathrm{SO}_4+8 \mathrm{HI} \longrightarrow \mathrm{H}_2 \mathrm{~S}+4 \mathrm{I}_2+4 \mathrm{H}_2 \mathrm{O} \\
\therefore x=8 \\
x=4 \\
z=4
\end{array}\)
Net increase in electron is '2' Sulphur is reduced, oxidation state changing from +6 to \(-2 \Rightarrow\) electrons
To balance 8 electrons, multiply \(H I\) and \(I_2\) by 4
Hence,
\(\begin{array}{ll}
\mathrm{H}_2 \mathrm{SO}_4+x \mathrm{HI} \longrightarrow \mathrm{H}_2 \mathrm{~S}+y \mathrm{I}_2+2 \mathrm{H}_2 \mathrm{O} \\
\Rightarrow \mathrm{H}_2 \mathrm{SO}_4+8 \mathrm{HI} \longrightarrow \mathrm{H}_2 \mathrm{~S}+4 \mathrm{I}_2+4 \mathrm{H}_2 \mathrm{O} \\
\therefore x=8 \\
x=4 \\
z=4
\end{array}\)
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