How many kilograms of NH3 are needed to produce 120 of pf3 if the reaction has 78.1% yield.
1 answer:
99.25 grams = .09925 Kg of fluorine gas is needed.
<h3>What are moles?</h3>A mole is defined as 6.02214076 × of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
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.
Hence, 99.25 grams = .09925 Kg of fluorine gas is needed.
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