How many grams of carbon tetrafluoride, CF_4, are produced from the complete reaction of 1.46 moles of carbon in the reaction C+2F_2 -> CF_4?

$n \left(\text{CF"_4)=n("C}\right) = 1.46 m o l$
$M \left({\text{CF}}_{4}\right) = \left(1 \times 12 + 4 \times 19\right) g m o {l}^{-} 1 = 88 g m o {l}^{-} 1$
$m \left({\text{CF}}_{4}\right) = n M = 1.46 m o l \times 88 g m o {l}^{-} 1 = 128 g$ (3s.f)