Phản ứng xảy ra:
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Ta có:
\(n_{Fe}=n_{H2}=\frac{1,68}{22,4}=0,075\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,075.56=4,2\left(g\right)\)
\(\Rightarrow m_{CuO}=14,2-4,2=10\left(g\right)\Rightarrow n_{CuO}=\frac{10}{64+16}=0,125\left(mol\right)\)
\(\Rightarrow\%m_{Fe}=\frac{4,2}{14,2}.100\%=29,58\%\)
\(\Rightarrow\%m_{CuO}=100\%-29,58\%=70,42\%\)
\(n_{HCl}=2n_{Fe}+2n_{CuO}=0,075.2+0,125.2=0,4\left(mol\right)\)
\(\Rightarrow CM_{HCl}=\frac{0,4}{0,1}=4M\)
B chứa FeCl2 0,075 mol và CuCl2 0,125 mol. Cho B tác dụng với NaOH dư.
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)
\(\Rightarrow n_{Fe\left(OH\right)_2}=n_{FeCl_2}=0,075\left(mol\right)\)
\(n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,125\left(mol\right)\)
\(\Rightarrow m_{kt}=0,075.\left(56+17.2\right)+0,125.\left(64+17.2\right)=19\left(g\right)\)
