Bài 1:
\(n_{H_2SO_4}=\dfrac{200.14,7\%}{98}=0,3\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\\ m_{Al_2\left(SO_4\right)_3}=342.0,1=34,2\left(g\right)\)
Bài 2:
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{Mg}=n_{H_2}=0,15\left(mol\right)\\ \Rightarrow m_{Mg}=0,15.24=3,6\left(g\right)\\ \%m_{Mg}=\dfrac{3,6}{10}.100=36\%\\ \%m_{Cu}=100\%-36\%=64\%\)