$n_{Mg} = \dfrac{8}{24} = \dfrac{1}{3}(mol)$
$Mg + 2HCl \to MgCl_2 + H_2$
$n_{HCl} = 2n_{Mg} = \dfrac{2}{3}(mol)$
$m_{HCl} = \dfrac{2}{3}.36,5 = 24,33(gam)$
\(n_{Mg}=\dfrac{8}{40}=0,2\left(mol\right)\)
Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,2 0,4
\(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
Chúc bạn học tốt
\(n_{Mg}=\dfrac{8}{24}=\dfrac{1}{3}\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{HCl}=\dfrac{1}{3}.2=\dfrac{2}{3}\left(mol\right)\\ m_{HCl}=36,5.\dfrac{2}{3}\approx24,333\left(g\right)\)
