a, PTHH: Mg + 2HCl → $MgCl_2$ + $H_2$ (1)
MgO + 2HCl → $MgCl_2$ + $H_2O$ (2)
`n_(H_2)` = `2.24/22.4` = 0.1 mol
Theo pt: `n_(Mg)` = `n_(H_2)` = 0.1 mol
→ `m_(Mg)` = 0.1 · 24 = 2.4 g → `m_(MgO)` = 6 - 2.4 = 3.6 g
→ `n_(MgO)` = `3.6/40` = 0.09 mol
%`m_(Mg)` = `2.4/6` · 100% = 40%
→ %`m_(MgO)` = 100% - 40% = 60%
b, Theo pt (1) `n_(HCl)` = 2`n_(H_2)` = 0.1 · 2 =0.2 mol
(2) `n_(HCl)` = 2`n_(MgO)` = 0.09 · 2 = 0.18 mol
→ `n_(HCl)` = 0.2 + 0.18 = 0.38 mol
$C_M$ = `0.38/0.2` = 1.9M
c, Theo pt: (1) `n_(MgCl_2)` = `n_(H_2)` = 0.1 mol
(2) `n_(MgCl_2)` = `n_(MgO)` = 0.09 mol
→ `n_(MgCl_2)` = 0.1 + 0.09 = 0.19 mol
→ `m_(MgCl_2)` = 0.19 · 95 = 18.05 g