When 1 mol of fluorine reacts with 2 mol of | vedclass

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(B) The balanced chemical equation for the reaction of fluorine with hot and concentrated $KOH$ is:$2F_2 + 4KOH 	o 4KF + 2H_2O + O_2$From the stoichi

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$2 : 1 : 0.5$

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(B) The balanced chemical equation for the reaction of fluorine with hot and concentrated $KOH$ is:$2F_2 + 4KOH 	o 4KF + 2H_2O + O_2$From the stoichiometry of the balanced equation,for $2 \ mol$ of $F_2$ and $4 \ mol$ of $KOH$,the products formed are $4 \ mol$ of $KF$,$2 \ mol$ of $H_2O$,and $1 \ mol$ of $O_2$.Dividing the coefficients by $2$ to match the $1 \ mol$ of $F_2$ given in the question:$1F_2 + 2KOH 	o 2KF + 1H_2O + 0.5O_2$Thus,the molar ratio of $KF : H_2O : O_2$ is $2 : 1 : 0.5$.

When $1 \ mol$ of fluorine reacts with $2 \ mol$ of hot and concentrated $KOH$,the products formed are $KF$,$H_2O$,and $O_2$. What is the molar ratio of $KF$,$H_2O$,and $O_2$ produced?

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