PTHH: Mg + 2HCl → MgCl2 + H2

- Đơn chất: Mg, H2

- Hợp chất: HCl, MgCl2

Ta có:

nH2 = 44,8/22,4 = 2mol

nHCl = 2nH2 = 2,2 = 4mol ⇒ mHCl = 4.(1 + 35,5) = 146gam

nMg = nH2 = 2mol ⇒ nMg = 2.24 = 48gam

PTHH: $\text{Mg + 2HCl $\longrightarrow$ $MgCl_{2}$ + $H_{2}$}$

Ta có: $\text{$n_{H_{2}}$ = $\frac{V}{22,4}$ = $\frac{44,8}{22,4}$ = 2 mol}$

Theo PTHH $\rightarrow$ $\text{$n_{HCl}$ = 2. $n_{H_{2}}$ = 4 mol}$

$\rightarrow$ $\text{$m_{HCl}$ = 4. 36,5 = 146 gam}$

Theo PTHH $\rightarrow$ $\text{$n_{Mg}$ = $n_{H_{2}}$ = 2 mol}$

$\rightarrow$ $\text{$m_{Mg}$ = 2. 24 = 48 gam}$

$\text{#Kudo}$