a) PTHH:
\(Fe\left(OH\right)_2+H_2SO_4\text{ loãng}\rightarrow FeSO_4+2H_2O\)
x → x
\(Cu\left(OH\right)_2+H_2SO_4\rightarrow CuSO_4+2H_2O\)
y → y
b) \(n_{H_2SO_4}=0,2\left(mol\right)\)
Đặt \(\left\{{}\begin{matrix}n_{Fe\left(OH\right)_2}=x\left(mol\right)\\n_{Cu\left(OH\right)_2}=y\left(mol\right)\end{matrix}\right.\)
Ta có:
\(\left\{{}\begin{matrix}BTKL:90x+98y=18,8\\BT\left(H_2SO_4\right):x+y=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(m_{Fe\left(OH\right)_2}=90\cdot0,1=9\left(g\right)\)
\(m_{Cu\left(OH\right)_2}=98\cdot0,1=9,8\left(g\right)\)
