Đáp án:

$\%m_{Al_2(SO_4)_3}=55,64\%$

Giải thích các bước giải:

 Bảo toàn điện tích:

$3.n_{Al^{3+}}+n_{Na^+}=n_{Cl^-}+2.n_{SO_4^{2-}}\\ \Rightarrow 0,3+a=b+0,22\\ \Rightarrow a-b=-0,08\ (1)\\ m_X=18,44\ (g)\\ \Rightarrow m_{Na^+}+m_{Al^{3+}}+m_{Cl^-}+m_{SO_4^{2-}}=18,44\ (g)\\ \Rightarrow 23a+2,7+35,5b+10,56=18,44\\ \Rightarrow 23a+35,5b=5,18\ (2)\\ \xrightarrow{\text{Từ (1),(2)}}\begin{cases}x=0,04\\y=0,12\end{cases}\\ \xrightarrow{\text{BTNT Na}}\ n_{Na_2SO_4}=\dfrac{1}{2}.n_{Na}=0,02(mol)\\ \xrightarrow{\text{BTNT S}}\ n_{SO_4^{2-}}=n_{Na_2SO_4}+3.n_{Al_2(SO_4)_3}\\ \Rightarrow 0,11=0,02+3..n_{Al_2(SO_4)_3}\\ \Rightarrow n_{Al_2(SO_4)_3}=0,03(mol)\\ \%m_{Al_2(SO_4)_3}=\dfrac{0,03.342}{18,44}.100\%=55,64\%$