\(Fe+2HCl\rightarrow FeCl2+H2\)(1)
1____2________1_______1
\(CuO+2HCl\rightarrow CuCl2+H2O\)(2)
1_______2__________1_____1
Ta có :\(n_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{Fe}=\frac{0,15.1}{1}=0,15\left(mol\right)\)
\(\rightarrow m_{Fe}=0,15.56=8,4g\)
\(\rightarrow m_{CuO}=10-8,4=1,6g\)
b,Theo PT(1)\(n_{HCl}=\frac{0,15.2}{1}=0,3\left(mol\right)\)
\(\rightarrow n_{CuO}=\frac{1,6}{80}=0,02\left(mol\right)\)
Theo PT(2)\(n_{HCl}=\frac{0,02.2}{1}=0,04\left(mol\right)\)
\(\rightarrow\Sigma n_{HCl}=0,3+0,04=0,34\left(mol\right)\)
\(\rightarrow CM_{HCl}=\frac{0,34}{0,2}=1,7M\)
