\(n_{H_{2}}=\dfrac{16,8}{22,4}=0,75(mol)\)
\(PTHH:Zn+2HCl\)→\(ZnCl_{2}+H_{2}\)↑
\(Theo \ PTHH:n_{Zn}=n_{H_{2}}=0,75 \ mol\\ =>m_{Zn}=65.0,75=48,75(g)\)
\(PTHH:2Al+6HCl\)→\(2AlCl_{3}+3H_{2}\)
\(Theo \ PTHH:n_{Al}=\dfrac{2}{3}n_{H_{2}}=\dfrac{2}{3}.0,75=0,5(mol)\\ =>m_{Al}=27.0,5=13,5(g)\)