nMg = 2.4/24 = 0.1 (mol)
Mg + 2HCl => MgCl2 + H2
0.1.......0.2..........0.1........0.1
mMgCl2 = 0.1*95 = 9.5 (g)
VH2 = 0.1*22.4 = 2.24 (l)
CM HCl = 0.2/0.4 = 0.5 (M)
Câu 17 :
\(a) Mg + 2HCl \to MgCl_2 + H_2\\ b) n_{MgCl_2} = n_{H_2} = n_{Mg} = \dfrac{2,4}{24} = 0,1(mol)\\ m_{MgCl_2} = 0,1.95 = 9,5(gam)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ c) n_{HCl} =2 n_{Mg} = 0,2(mol)\\ C_{M_{HCl}} = \dfrac{0,2}{0,4} = 0,5M\)