write the equation for the reaction
that is 6 F2 +P4 =4 PF3
find the theoretical mass that is
let the theoretical yield be represented by y
theoretical yield = 78.1/100 = 120/y
y= 153.6 grams
find the number of moles of PF3
moles = mass/molar mass
= 153.6/87.97 =1.746 moles
by use of mole ratio between F2 :PF3 which is 6:4 the moles of F2 is therefore= 1.746 x 6/4 = 2.62 moles
mass = moles x molar mass
= 1.746 moles x38 g/mol = 99.6 grams
The mass of F₂ needed to produce 120 g of PF₃ if the reaction has a 78.1% yield is 99.49 g
We'll begin by calculating the theoretical yield of PF₃. This can be obtained as follow:
Actual yield of PF₃ = 120 g
Percentage yield of PF₃ = 78.1% = 0.781
[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\0.781 = \frac{120}{Theoretical}\\\\[/tex]
Cross multiply
0.781 × Theoretical = 120
Divide both side by 0.781
[tex]Theoretical = \frac{120}{0.781}\\\\[/tex]
Theoretical yield of PF₃ = 153.6 g
Next, we shall determine the mass of F₂ that reacted from the balanced equation. This is illustrated below:
6F₂ + P₄ —> 4PF₃
Molar mass of F₂ = 2 × 19 = 38 g/mol
Mass of F₂ from the balanced equation = 6 × 38 = 228 g
Molar mass of PF₃ = 31 + (3×19) = 88 g/mol
Mass of PF₃ from the balanced equation = 4 × 88 = 352 g
From the balanced equation above,
228 g of F₂ reacted to produce 352 g of PF₃.
Finally, we shall determine the mass of F₂ needed to produce 153.6 g of PF₃. This can be obtained as follow:
From the balanced equation above,
228 g of F₂ reacted to produce 352 g of PF₃.
Therefore,
Xg of F₂ will react to produce 153.6 g of PF₃ i.e
Xg of F₂ = [tex]\frac{228 * 153.6}{352}\\[/tex]
Thus, the mass of F₂ needed to produce 120 g of PF₃ if the reaction has a 78.1% yield is 99.49 g
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