Đáp án + Giải thích các bước giải:

$a\big)$

$n_{Al}=\dfrac{2,7}{27}=0,1(mol)$

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

Theo PT: $n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=0,15(mol)$

$\to m_{dd\,H_2SO_4}=\dfrac{0,15.98}{9,8\%}=150(g)$

$b\big)$

Theo PT: $\begin{cases} n_{H_2}=n_{H_2SO_4}=0,15(mol)\\ n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=0,05(mol) \end{cases}$

$\to C\%_{Al_2(SO_4)_3}=\dfrac{0,05.342}{2,7+150-0,15.2}.100\%\approx 11,22\%$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{2,7}}{{27}} = 0,1mol\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,15mol\\
{n_{{H_2}S{O_4}}} = \frac{3}{2}{n_{Al}} = 0,15mol\\
{m_{{H_2}S{O_4}}} = 0,15 \times 98 = 14,7g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{14,7 \times 100}}{{9,8}} = 150g\\
b)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,05mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,05 \times 342 = 17,1g\\
C{\% _{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{17,1}}{{2,7 + 150 - 0,3}} \times 100\%  = 11,22\% 
\end{array}\)