You are given 90.0 grams of fluorine with excess aluminum. How many grams will you produce

Answer 1

2Al(s) +3F₂(g) ===> 2AlF₃(s)

moles F2(g) present = 90.0 g x 1 mol/37.98 g = 2.370 moles F2(g)

mass AlF₃ produced = 2.37 moles F₂ x 2 moles AlF₃/3 moles F₂ x 83.98 g/mole = 133 g AlF₃ (to 3 sig. figs.)