Đáp án:

a) K2SO4: 50,44%; Al2(SO4)3: 49,56%

b) 105,68 g

Giải thích các bước giải:

 a) Gọi x, y là số mol $A{l_2}{(S{O_4})_3};{K_2}S{O_4}$

tỉ lệ số nguyên tử O = $\dfrac{{12x + 4y}}{{17x + 7y}} = \dfrac{{20}}{{31}}$

$\begin{gathered}
   \Leftrightarrow 372x + 124y = 340x + 140y \hfill \\
   \Rightarrow 32x = 16y \Rightarrow y = 2x \hfill \\ 
\end{gathered} $

$\% {m_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{342x}}{{342x + 174y}}.100\%  = \dfrac{{342x}}{{342x + 174.2x}}.100\%  = 49,56\% $

$\% {m_{{K_2}S{O_4}}} = 100 - 49,56 = 50,44\% $

b) 

$\begin{gathered}
  {m_{{K_2}S{O_4}}} = {m_X}.50,44\%  = 82,8.50,44\%  = 41,76\left( g \right) \hfill \\
   \Rightarrow {m_{A{l_2}{{(S{O_4})}_3}}} = 82,8 - 41,76 = 41,04\left( g \right) \hfill \\
  {n_{{K_2}S{O_4}}} = \dfrac{{41,76}}{{174}} = 0,24mol;{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{41,04}}{{342}} = 0,12mol \hfill \\
  {n_{Ba{{(OH)}_2}}} = 0,4mol \hfill \\ 
\end{gathered} $

$\begin{gathered} 3Ba{(OH)_2} + A{l_2}{(S{O_4})_3} \to 2Al{(OH)_3} \downarrow + 3BaS{O_4} \downarrow \hfill \\ {\text{0,36 }} \leftarrow {\text{ 0,12 }} \to {\text{ 0,24 0,36}} \hfill \\ {\text{Ba(OH}}{{\text{)}}_2} + 2Al{(OH)_3} \to Ba{(Al{O_2})_2} + 4{H_2}O \hfill \\ {\text{0,04 }} \to {\text{ 0,08 0,04}} \hfill \\ Ba{(Al{O_2})_2} + {K_2}S{O_4} \to BaS{O_4} \downarrow + 2KAl{O_2} \hfill \\ {\text{0,04 }} \to {\text{ 0,04 0,04}} \hfill \\ \end{gathered} $

Kết tủa gồm $A{l_2}{(S{O_4})_3}(0,16{\text{ mol); }}BaS{O_4}(0,4{\text{ mol)}}$

$m = {m_{A{l_2}{{(S{O_4})}_3}}} + {m_{BaS{O_4}}} = 0,16.78 + 0,4.233 = 105,68\left( g \right)$