Gọi số mol của KOH là x.
\(2KOH\left(x\right)+CuSO_4\left(\dfrac{x}{2}\right)\rightarrow K_2SO_4\left(\dfrac{x}{2}\right)+Cu\left(OH\right)_2\left(\dfrac{x}{2}\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{KOH}=56x\\m_{CuSO_4}=80x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ddKOH}=\dfrac{56x}{0,112}=500x\left(g\right)\\m_{ddCuSO_4}=\dfrac{80x}{0,16}=500x\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_4}=87x\\m_{Cu\left(OH\right)_2}=49x\end{matrix}\right.\)
\(\Rightarrow\%K_2SO_4=\dfrac{87x}{500x+500x-49x}.100\%=9,14\%\)