Bài 1 :
\(n_{CuO}=\dfrac{4.8}{80}=0.06\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(0.06.....0.06...0.06\)
\(V_{H_2}=0.06\cdot22.4=1.344\left(l\right)\)
\(m_{Cu}=0.06\cdot64=3.84\left(g\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.06...................................0.06\)
\(m_{Fe}=0.06\cdot56=3.36\left(g\right)\)
Bài 2 :
\(n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(\dfrac{1}{30}.......0.1...........0.05\)
\(m_{Al}=\dfrac{1}{30}\cdot27=0.9\left(g\right)\)
\(m_{HCl\left(dư\right)}=\left(0.2-0.1\right)\cdot36.5=3.65\left(g\right)\)
