A reaction mixture contained  $100.0g$ of  $K_{2}Mn{F}_6$  and  $174g$ of  $Sb{F}_5.$ Fluorine was produced in 40 % yield. What is the mass of  $F_2$  produced in grams.
 $2K_{2}Mn{F}_6+4Sb{F}_5\to 4KSb{F}_6+2Mn{F}_3+F_2$
At wts : K = 39, Mn = 55, F = 19, Sb = 122

$\begin{array}{l}2K_{2}Mn{F}_6+4Sb{F}_5\to 4KSb{F}_6+2Mn{F}_3+F_2\\ 494 8681100224 38\end{array}$

$494K_{2}Mn{F}_6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}require\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}868gofSb{F}_5$

$100g{\text{\hspace{0.17em}of\hspace{0.17em}require - 175.7 SbF}}_5$

Since the amount taken is $174gofSb{F}_5,$  it will be limiting reagent and it can react with  
$$\begin{gathered} 99gof{K}_{2}Mn{F}_6 \\ 99gm{K}_{2}Mn{F}_6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}produce\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}7.617g\text{\hspace{0.17em}of\hspace{0.17em}}{F}_2 \end{gathered}$$
Since yields is 40% the amount of  $F_2$ produced is
$\begin{array}{l}100\%\_\_\_\_\_7.617\\ 40\%\_\_\_\_\_?\end{array}$

=3.04-3.05 gm (range may be given)