a) Gọi \(\left\{{}\begin{matrix}n_{CuSO_4}=a\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=b\left(mol\right)\end{matrix}\right.\)
\(n_{NaOH}=0,5.2,4=1,2\left(mol\right)\)
PTHH: \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
a---------->2a----------->a
\(Fe_2\left(SO_4\right)_3+6NaOH\rightarrow2Fe\left(OH\right)_3\downarrow+3Na_2SO_4\)
b---------------->6a----------->2b
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
a------------->a
\(2Fe\left(OH\right)_3\xrightarrow[]{t^o}Fe_2O_3+3H_2O\)
2b------------>b
=> \(\left\{{}\begin{matrix}2a+6b=1,2\\80a+160b=40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C_{M\left(CuSO_4\right)}=\dfrac{0,3}{0,1}=3M\\C_{M\left(Fe_2\left(SO_4\right)_3\right)}=\dfrac{0,1}{0,1}=1M\end{matrix}\right.\)
b) \(CuSO_4+Ba\left(NO_3\right)_2\rightarrow BaSO_4\downarrow+Cu\left(NO_3\right)_2\)
0,3------------------------>0,3
\(Fe_2\left(SO_4\right)_3+3Ba\left(NO_3\right)_2\rightarrow3BaSO_4\downarrow+2Fe\left(NO_3\right)_3\)
0,1------------------------------->0,3
=> \(m_{kt}=m_{BaSO_4}=\left(0,3+0,3\right).233=139,8\left(g\right)\)
