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The number of moles of oxygen obtained by the electrolytic decomposition of 108 g water is
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\(\begin{array}{ll}
2 \mathrm{H}_2 \mathrm{O} \xrightarrow{\text { Electrolysis }} 2 \mathrm{H}_2+ &\mathrm{O}_2 \\
2 \text{mol} & 1 \text{mol} \\
2 \times 18 =36 g
\end{array}\)
\(\because 36 \mathrm{~g}\) of \(\mathrm{H}_2 \mathrm{O}\) produce 1 mole of oxygen
\(\therefore 108 \mathrm{~g}\) of water will produce oxygen
\(=\frac{108}{36}=3 \mathrm{~mol}\)
2 \mathrm{H}_2 \mathrm{O} \xrightarrow{\text { Electrolysis }} 2 \mathrm{H}_2+ &\mathrm{O}_2 \\
2 \text{mol} & 1 \text{mol} \\
2 \times 18 =36 g
\end{array}\)
\(\because 36 \mathrm{~g}\) of \(\mathrm{H}_2 \mathrm{O}\) produce 1 mole of oxygen
\(\therefore 108 \mathrm{~g}\) of water will produce oxygen
\(=\frac{108}{36}=3 \mathrm{~mol}\)
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