Bài 19:
\(a,PTHH:Mg+2HCl\to MgCl_2+H_2\\ \text {Đơn chất: }Mg,H_2\\ \text {Hợp chất: }HCl,MgCl_2\\ b,n_{H_2}=\dfrac{44,8}{22,4}=2(mol)\\ \Rightarrow n_{HCl}=2n_{H_2}=4(mol);n_{Mg}=n_{H_2}=2(mol)\\ \Rightarrow m_{HCl}=4.36,5=146(g);m_{Mg}=2.24=48(g)\)