Question

Given the following equation, if I make 18.5G of iron (III) sulphate, what is the percent yield? I picked B but I’m not sure if I’m correct

Answer 1

B. The percent yield is 55.9%.

Step-by-step explanation:

From the balanced reaction we know that 2 moles of FePO4 produce 1 mole of Fe2(SO4)3.

1st) It necessary to convert moles into grams, using the molar mass of FePO4 (151g/mol) and the molar mass of Fe2(SO4)3 (400g/mol):

- FePO4 conversion:

`2moles*(151g)/(1mole)=302g`

- Fe2(SO4)3 conversion:

In this case, it is not necessary the conversion, because there is 1 mole, so it is equal to 400g.

Now we know that 302g of FePO4 produce 400g of Fe2(SO4)3.

2nd) Now we have to calculate the Theoretical yield of Iron (III) sulfate, that is, the amount of Iron (III) sulfate that we must produce from 25g of Iron (III) phosphate:

`\begin{gathered} 302gFePO_4-400gFe_2(SO_4)_3 \\ 25gFePO_4-x=(25gFePO_4*400gFe_2(SO_4)_3)/(302gFePO_4) \\ x=33.11gFe_2(SO_4)_3 \end{gathered}`

So, the Theoretical yield is 33.11g.

3rd) Finally, we can calculate the Percent yield using the Theoretical yield (33.11g) and the Actual yield (18.5g):

`\begin{gathered} PercentYield=(ActualYield)/(TheoreticalYield)*100\% \\ PercentYield=(18.5g)/(33.11g)*100\% \\ PercentYield=55.9\% \end{gathered}`

So, the percent yield is 55.9%.