Answer
Please see the work below.
Work Step by Step
We know that
$10.0g BF_3\times \frac{1 mol \space BF_3}{67.81g\space g\space BF_3}\times \frac{1\space mol \space BF_3}{1 mol\space BF_3}\times \frac{84.84g BF_3}{1 \space mol \space BF_3}=12.5g$