Consider the reaction of iron (III) sulfate (aq) and sodium phosphate (aq) to fo

Question

Consider the reaction of iron (III) sulfate (aq) and sodium phosphate (aq) to form iron (III) phosphate (s) and sodium sulfate (aq).

c. If 2.16 g of iron (III) phosphate are obtained when a student performed this reaction, what is the percent yield?

d. Is the answer from part C reasonable? Assuming no calculation errors were made, give a possible explanation for the answer in part C.

e. If you perform this reaction with 20.00 mL of the 0.250 M solution of iron (III) sulfate and obtain a 65.0% yield, how many grams of iron (III) phosphate will you make?

Explanation / Answer

a)

the balanced reaction is

Fe2(s04)3 + 2Na3P04   ----> 2FeP04 + 3Na2S04

b)

we know that

moles = molarity x volume (L)

so

moles of Fe2(s04)3 = 0.25 x 25 x 10-3

moles of Fe2(s04)3 = 6.25 x 10-3

from the above reaction

moles of FeP04 formed = 2 x moles of Fe2(S04)3 reacted

so

moles of FeP04 formed = 2 x 6.25 x 10-3

moles of FeP04 formed = 12.5 x 10-3

now

mass = moles x molar mass

so

mass of FeP04 formed = 12.5 x 10-3 x 150.82

mass of FeP04 formed = 1.885 g

so

1.885 g of FeP04 can be made

c)

percent yield = ( actual / theoretical ) x 100

percent yield = ( 2.16 / 1.885 ) x 100

percent yield = 114.58

the percent yield is 114.58%

d)

part C is not reasonable , because percent yield cannot be greater than 100

that is we cannot produce the amount greater than the theoretical value

e)

we know that

moles = molarity x volume (L)

so

moles of Fe2(s04)3 = 0.25 x 20 x 10-3

moles of Fe2(s04)3 = 5 x 10-3

from the above reaction

moles of FeP04 formed = 2 x moles of Fe2(S04)3 reacted

so

moles of FeP04 formed = 2 x 5 x 10-3

moles of FeP04 formed = 10 x 10-3

now

mass = moles x molar mass

so

mass of FeP04 formed = 10 x 10-3 x 150.82

theoretical mass of FeP04 formed = 1.5082 g

given

percent yield = 65

so

( actual / theoretical ) x 100 = 65

actual / 1.5082 = 0.65

actual = 0.98 g

so

0.98 g of FeP04 can be made

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