Question

Consider the following 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?

Answer 1

Answer:

44.28 grams.

Explanation:

Let us write the balanced reaction:

$P_{4}+6F_{2}-->4PF_{3}$

As per balanced equation, six moles of fluorine gas will give four moles of PF₃.

The mass of PF₃ required = 120 g

The molar mass of PF₃ = 88g/mol

Moles of PF₃ required =$\frac{mass}{molarmass}=\frac{120}{88}=1.364mol$

The moles of fluorine gas required = $\frac{4X1.364}{6}=0.91$

the mass of fluorine gas required = moles X molar mass = 0.91x38 = 34.58g

Now this much mass will be required if the reaction is of 100% yield

But as given that the yield of reaction is only 78.1%

The mass of fluorine required = $\frac{massX100}{78.1} =\frac{34.58X100}{78.1} =44.28g$