a)Mg + 2HCl -> MgCl_2 + H_2

n_{Mg} = {1,2}/{24} = 0,05 mol

50ml = 0,05l

n_{HCl} = 0,05 . 3 = 0,15 mol

Vì 0,05 < 0,15 => HCl dư

b)n_{H_2} = n_{Mg} = 0,05 mol

=> V_{H_2} = 0,05 . 22,4 = 1,12l

c)n_{MgCl_2} = n_{Mg} = 0,05 mol

=> m_{MgCl_2} =0,05 . 95 = 4,75g

n_{HCl p//ứ} = 2n_{Mg} = 2 . 0,05 = 0,1 mol

d)n_{HCl dư} = 0,15 - 0,1= 0,05 mol

=> C_{M_{HCl dư}} = {0,05}/{0,05} = 1M

C_{M_{MgCl_2}} = {0,05}/{0,05} = 1M

a)\ Mg+2HCl->MgCl_2+H_2

b)

50mL=0,05L

n_{Mg}={1,2}/{24}=0,05(mol)

n_{HCl}=0,05.3=0,15(mol)

Do 0,05<{0,15}/2->HCl dư.

Theo PT: n_{H_2}=n_{Mg}=0,05(mol)

->V_{H_2}=0,05.22,4=1,12(L)

c)

Theo PT: n_{MgCl_2}=n_{Mg}=0,05(mol)

->m_{MgCl_2}=0,05.95=4,75(g)

d)

Theo PT: n_{HCl\ pu}=2n_{Mg}=0,1(mol)

->C_{M\ HCl\ du}={0,15-0,1}/{0,05}=1M

        C_{M\ MgCl_2}={0,05}/{0,05}=1M