A reaction mixture contained $100.0 g$ of $K_{2} M n F_{6}$ and $174 g$ of $S b F_{5} .$ Fluorine was produced in 40 % yield. What is the mass of $F_{2}$ produced in grams.
$2 K_{2} M n F_{6} + 4 S b F_{5} \to 4 K S b F_{6} + 2 M n F_{3} + F_{2}$
At wts : K = 39, Mn = 55, F = 19, Sb = 122

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Detailed Solution

$\begin{array}{l} 2 K_{2} M n F_{6} + 4 S b F_{5} \to 4 K S b F_{6} + 2 M n F_{3} + F_{2} \\ 494 868 1100 224 38 \end{array}$

$494 K_{2} M n F_{6} require 868 g o f S b F_{5}$

$100 g {of require - 175.7 SbF}_{5}$

Since the amount taken is $174 g o f S b F_{5},$ it will be limiting reagent and it can react with
$99 g o f K_{2} M n F_{6} 99 g m K_{2} M n F_{6} produce 7.617 g of F_{2}$
Since yields is 40% the amount of $F_{2}$ produced is
$\begin{array}{l} 100 %_____7.617 \\ 40 %_____? \end{array}$