Câu 1:
a, PT: \(Fe\left(OH\right)_2+H_2SO_4\rightarrow FeSO_4+2H_2O\)
\(Cu\left(OH\right)_2+H_2SO_4\rightarrow CuSO_4+2H_2O\)
b, Gọi: \(\left\{{}\begin{matrix}n_{Fe\left(OH\right)_2}=x\left(mol\right)\\n_{Cu\left(OH\right)_2}=y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow90x+98y=18,8\left(1\right)\)
Ta có: \(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{Fe\left(OH\right)_2}+n_{Cu\left(OH\right)_2}=x+y=0,2\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe\left(OH\right)_2}=0,1.90=9\left(g\right)\\m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\end{matrix}\right.\)
