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One mole of fluorine is reacted with two moles of hot concentrated $\mathrm{KOH}$. The products formed are $\mathrm{KF}, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{O}_2$. The molar ratio of $\mathrm{KF}, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{O}_2$, respectively is
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Verified Answer
The correct answer is:
$2: 1: 0.5$
Stoichiometric equation for the reaction is
$$
\mathrm{F}_2+2 \mathrm{KOH} \longrightarrow 2 \mathrm{KF}+\mathrm{H}_2 \mathrm{O}+\frac{1}{2} \mathrm{O}_2
$$
$$
\mathrm{F}_2+2 \mathrm{KOH} \longrightarrow 2 \mathrm{KF}+\mathrm{H}_2 \mathrm{O}+\frac{1}{2} \mathrm{O}_2
$$
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