a)\(2Al+6HCl-->2AlCl3+3H2\)
\(CuO+2HCl-->CuCl2+H2O\)
\(n_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{Al}=\frac{2}{3}n_{H2}=0,1\left(mol\right)\)
\(m_{Al}=0,1.27=2,7\left(g\right)\)
\(\%m_{Al}=\frac{2,7}{12,3}.100\%=21,95\%\)
\(\%m_{CuO}=100-21,95=78,05\%\)
b)\(n_{HCl\left(1\right)}=2n_{H2}=0,3\left(mol\right)\)
\(m_{CuO}=12,3-2,7=9,6\left(g\right)\)
\(n_{CuO}=\frac{9,6}{80}=0,12\left(mol\right)\)
\(n_{HCl\left(2\right)}=2n_{CuO}=0,24\left(mol\right)\)
\(\sum n_{HCl}=0,3+0,24=0,54\left(mol\right)\)
\(V_{HCl}=\frac{0,54}{1,5}=0,3\left(l\right)\)