# Calculating yield percentage of magnetite synthesis

I’ve made magnetite $$\ce{Fe3O4}$$ in the lab and now I need to calculate yield percentage. How can I do it? I have these data.

\begin{align} \ce{FeSO4.7H2O(aq) &-> Fe^2+(aq) + SO4^2- (aq) + 7H2O(l)}\\ \ce{KOH(aq) &-> K+(aq) + OH- (aq)}\\ \ce{KNO3(aq) &-> K+(aq) + NO3- (aq)}\\ \ce{Fe^2+(aq) + 2OH- (aq) &-> Fe(OH)2(s)}\\ \ce{3Fe(OH)2(s) + NO3- (aq) &-> Fe3O4(s) + NO2- (aq) + 3H2O(l)}\\ \end{align}

\begin{align} \ce{FeSO4.7H2O &->} 4.20~\mathrm{g}\\ \ce{KOH &->} 1.90~\mathrm{g}\\ \ce{KNO3 &->} 0.12~\mathrm{g}\\ \text{Magnetite}~\ce{&->} 0.36~\mathrm{g}\\ \end{align}

I know how to calculate the yield for simple reactions but I don’t know how to do it when there are many reactions involved.

• Ok thanks, as I said I know how to calculate yield for simple reactions but not for multistep synthesis. (ac) state symbol is the same as (aq) except that in spanish it is common to write (ac) ("aqueous" = "acuosa"), english is not my native language sorry. – Ragnar Jun 18 '15 at 0:30
• then share the result with us -post an answer – Jaroslav Kotowski Jun 18 '15 at 9:20

$$\ce{3FeSO4.7H2O(aq) + 6KOH(aq) + KNO3(aq) -> Fe3O4(s) + NO2-(aq) + ...}$$