Answer is: 99.25 grams of fluorine gas is needed.
Balanced chemical reaction: P₄ (s) + 6F₂ (g) → 4PF₃ (g).
m(PF₃) = 120.0 g; mass of phosphorus trifluoride.
n(PF₃) = m(PF₃) ÷ M(PF₃).
n(PF₃) = 120 g ÷ 88 g/mol.
n(PF₃) = 1.36 mol.
From chemical reaction: n(PF₃) : n(F₂) = 4 : 6 (2 : 3).
n(F₂) = 1.36 mol · 3 / 2.
n(F₂) = 2.04 mol; amount of substance.
m(F₂) = n(F₂) · M(F₂).
m(F₂) = 2.04 mol · 38 g/mol.
m(F₂) = 77.52 g ÷ 0.781.
m(F₂) = 99.25 g.
