\(n_{HCl}=0.1\cdot2=0.2\left(mol\right)\)
\(n_{H_2SO_4}=0.1\cdot1.5=0.15\left(mol\right)\)
\(n_{H^+}=0.2+0.15\cdot2=0.5\left(mol\right)=2n_{H_2}\)
\(Mg+2H^+\rightarrow Mg^{2+}+H_2\)
\(2Al+6H^+\rightarrow2Al^{3+}+3H_2\)
Bảo toàn khối lượng :
\(m_{Muối}=5.1+0.2\cdot36.5+0.15\cdot98-0.25\cdot2=26.6\left(g\right)\)