Answer
Please see the work below.
Work Step by Step
We know that
$\%F=\frac{mass\space of F\space in \space 1mol\space SnF_2}{molar\space mass\space of SnF_2}\times 100\%=\frac{2(19.00g)}{156.7g}\times \%=24.25\%$
Now: $(0.2425)(24.6)=5.97g\space F$
