Question
One mole of fluorine is reacted with two moles of hot and concentrated $$KOH.$$ The products formed are $$KF,{H_2}O$$ and $${O_2}.$$ The molar ratio of $$KF,{H_2}O$$ and $${O_2}$$ respectively is
A.
1 : 1 : 2
B.
2 : 1 : 0.5
C.
1 : 2 : 1
D.
2 : 1 : 2
Answer :
2 : 1 : 0.5
Solution :
$$2{F_2} + 4KOH \to 4KF + {O_2} + 2{H_2}O$$ for 1 $$mole$$ of $${F_2}$$ the molar ratio.
$$\eqalign{
& {F_2}\,\,\,\,\,\,KOH\,\,\,\,\,\,KF\,\,\,\,\,\,{O_2}\,\,\,\,\,\,{H_2}O \cr
& 1\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,1 \cr} $$