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?

Answer 1

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