1 November, 16:51

# Consider the unbalanced reaction: p4 (s) + f2 (g) → pf3 (g) what mass of fluorine gas is needed to produce 120. g of pf3 if the reaction has a 78.1 % yield?

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1. 1 November, 16:52
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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.
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