Question

If the percent yield of Fe2(SO4)3 in the reaction was 67% what was the actual yield? Equation is Fe+H2SO4 Fe2(SO4)3+H2

Answer 1

Answer:

The actual yield is 268g.

Explanation:

1st) It is necessary to balance the chemical reaction:

$2Fe+3H_2SO_4\rightarrow Fe_2(SO_4)_3+3H_2$

From the balanced reaction we know that 1 moles of Fe2(SO4)3 is formed from 2 moles of Fe and 3 moles of H2SO4.

2nd) With the molar mass of Fe2(SO4)3, we have convert the mole to grams:

- Fe2(SO4)3 molar mass: 400g/mol

So, 1 mole of Fe2(SO4)3 is equal to 400g. This will be the Theroretical yield.

3rd) Now, with the Theoretical yield and the Percent yield, we can calculate the Actual yield of Fe(SO4)3:

$\begin{gathered} PercentYield=\frac{ActualYield}{TheoreticalYield}*100\% \\ 67\%=\frac{ActualY\imaginaryI eld}{400g}\times100 \\ \frac{67\%}{100\%}*400g=ActualY\mathrm{i}eld \\ 268g=ActualY\mathrm{i}eld \end{gathered}$

Finally, the actual yield is 268g.