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

\begin{align}
\ce{FeSO4.7H2O &amp;-&gt;} 4.20~\mathrm{g}\\  
\ce{KOH &amp;-&gt;} 1.90~\mathrm{g}\\  
\ce{KNO3 &amp;-&gt;} 0.12~\mathrm{g}\\  
\text{Magnetite}~\ce{&amp;-&gt;} 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:

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

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.