The amount of water produced (in g ) in the oxidation of 1 mole of rhombic sulphur by conc. $\mathrm{HNO}_3$ to a compound with the highest oxidation state of sulphur is $\_\_\_\_$
(Given data : Molar mass of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}$ )
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Solution
\begin{aligned}
&\mathrm{S}_8+48 \mathrm{HNO}_3 \longrightarrow 8 \mathrm{H}_2 \mathrm{SO}_4+48 \mathrm{NO}_2+16 \mathrm{H}_2 \mathrm{O}\\
&1 \text { mole of rhombic sulphur produce } 16 \text { mole of } \mathrm{H}_2 \mathrm{O} \text { i.e. } 288 \mathrm{gm} \text { of } \mathrm{H}_2 \mathrm{O}
\end{aligned}
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