Đáp án đúng: C
15 gam.
$\begin{array}{l}\text{M+}{{\text{H}}_{\text{2}}}\text{O}\xrightarrow{{}}\text{MOH+}\frac{\text{1}}{\text{2}}{{\text{H}}_{\text{2}}}\\\text{MOH+Al+}{{\text{H}}_{\text{2}}}\text{O}\xrightarrow{{}}\text{MAl}{{\text{O}}_{\text{2}}}\text{+}\frac{\text{3}}{\text{2}}{{\text{H}}_{\text{2}}}\end{array}$Do còn chất rắn không tan ⇒ Al dư$\Rightarrow {{\text{n}}_{\text{M}}}\text{=}{{\text{n}}_{\text{MAl}{{\text{O}}_{\text{2}}}}}\text{=}\frac{\text{1}}{\text{2}}{{\text{n}}_{{{\text{H}}_{\text{2}}}}}\text{=0}\text{,161}\ \text{mol}$
$\displaystyle \text{MAl}{{\text{O}}_{\text{2}}}\text{ }\!\!~\!\!\text{ + HCl}~\to \text{ }\!\!~\!\!\text{ MCl + Al}{{\left( \text{OH} \right)}_{\text{3}}}$nHCl = x > 0,18 mol $\Rightarrow {{\text{n}}_{\text{Al(OH}{{\text{)}}_{\text{3}}}}}\text{=}\frac{\text{1}}{\text{3}}\text{(4}{{\text{n}}_{\text{Al}{{\text{O}}_{\text{2}}}}}\text{-}\,{{\text{n}}_{{{\text{H}}^{\text{+}}}}}\text{)=}\frac{\text{(0}\text{,644-x)}}{\text{3}}$Và${{\text{n}}_{\text{AlC}{{\text{l}}_{\text{3}}}}}\text{=}\frac{\text{1}}{\text{3}}\text{(}{{\text{n}}_{\text{HCl}}}\text{-}\,{{\text{n}}_{\text{Al}{{\text{O}}_{\text{2}}}}}\text{)=}\frac{\text{(x}\,\text{-}\,\text{0}\text{,161)}}{\text{3}}\ \text{mol;}\ {{\text{n}}_{\text{MCl}}}\text{=}\,{{\text{n}}_{\text{MAl}{{\text{O}}_{\text{2}}}}}\text{=}\,\text{0}\text{,161}\ \text{mol}$$\begin{array}{l}\Rightarrow {{\text{m}}_{\text{AlC}{{\text{l}}_{\text{3}}}}}\text{+}{{\text{m}}_{\text{MCl}}}\text{=11}\text{,9945}\ \text{gam}\,\text{=}\,\text{44}\text{,5(x}\,\text{-}\,\text{0}\text{,161) +}\,\text{0}\text{,161}\text{.(M+35}\text{,5)}\\\Rightarrow ~\text{44}\text{,5x + 0}\text{,161M = 13}\text{,4435}\end{array}$Do x > 0,18 ⇒ M < 33,75 ⇒ M là Na (Li không tan nhiều trong nước)⇒ nNa = 0,161 và nAl pứ = 0,161⇒ m – 0,4687m = 0,161.23 + 0,161.27 ⇒ m = 15,15 gam
