One mole of fluorine is reacted with two moles of hot concentrated $\mathrm{KOH}$. The products formed are…
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
$1: 1: 2$
$2: 1: 0.5$
$1: 2: 1$
$2: 1: 2$
Solution
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
$