Ta có: \(n_{CuSO_4.5H_2O}=\dfrac{a}{250}\left(mol\right)=n_{CuSO_4\left(1\right)}\)
\(\Rightarrow m_{CuSO_4\left(1\right)}=\dfrac{a}{250}.160=\dfrac{16a}{25}\left(g\right)\)
\(m_{CuSO_4\left(2\right)}=b.c\%=\dfrac{bc}{100}\left(g\right)\)
\(\Rightarrow\Sigma m_{CuSO_4}=\dfrac{16}{25}a+\dfrac{bc}{100}\left(g\right)\)
Mà: \(C\%_{CuSO_4}=d\%\) \(\Rightarrow\dfrac{\dfrac{16a}{25}+\dfrac{bc}{100}}{a+b}.100\%=d\%\)