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.




Add up the reactions (double the second one so there is sufficient hydroxide) to get a net reaction with your starting materials as reactants and your desired product as one of the products:


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


Then, find the limiting reactant, calculate the theoretical yield from it, and compare it to the amount of product to figure out the percent yield.


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