$n_{Al_2O_3}=\dfrac{30,6}{102}=0,3(mol);n_{H_2SO_4}=2.0,6=1,2(mol)\\ Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ LTL:\dfrac{0,3}{1}<\dfrac{1,2}{3}\Rightarrow H_2SO_4\text{ dư}\\ \Rightarrow n_{Al_2(SO_4)_3}=n_{Al_2O_3}=0,3(mol);n_{H_2SO_4(dư)}1,2-0,3.3=0,3(mol)\\ a,C_{M_{Al_2(SO_4)_3}}=\dfrac{0,3}{0,6}=0,5M\\ C_{M_{H_2SO_4}}=\dfrac{0,3}{0,6}=0,5M\\ b,2Al_2O_3\xrightarrow[criolit]{đpnc}4Al+3O_2\\ \Rightarrow n_{Al}=2n_{Al_2O_3}=0,6(mol)\\ \Rightarrow m_{Al}=0,6.27=16,2(g)\\ \Rightarrow m_{Al(tt)}=16,2.95\%=15,39(g)$

Đáp án:

\(\begin{array}{l}
a)\\
{C_M}A{l_2}{(S{O_4})_3} = {C_M}{H_2}S{O_4} = 0,5M\\
b){m_{Al}} = 15,39g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{A{l_2}{O_3}}} = \dfrac{{30,6}}{{102}} = 0,3\,mol\\
{n_{{H_2}S{O_4}}} = 0,6 \times 2 = 1,2\,mol\\
{n_{A{l_2}{O_3}}} < \dfrac{{{n_{{H_2}S{O_4}}}}}{3} \Rightarrow {H_2}S{O_4} \text{ dư }\\
{n_{A{l_2}{{(S{O_4})}_3}}} = {n_{A{l_2}{O_3}}} = 0,3\,mol\\
{n_{{H_2}S{O_4}}} \text{ dư }= 1,2 - 0,3 \times 3 = 0,3\,mol\\
{C_M}A{l_2}{(S{O_4})_3} = {C_M}{H_2}S{O_4} \text{ dư } = \dfrac{{0,3}}{{0,6}} = 0,5M\\
b)\\
2A{l_2}{O_3} \xrightarrow{\text{ điện phân nóng chảy }} 4Al + 3{O_2}\\
{n_{Al}} = 2{n_{A{l_2}{O_3}}} = 0,6\,mol\\
H = 95\%  \Rightarrow {m_{Al}} = 0,6 \times 27 \times 95\%  = 15,39g
\end{array}\)