Đáp án:
a) 6,3g
b) 11,4g
c) 2,24l
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{8,1}}{{27}} = 0,3\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{9,8}}{{98}} = 0,1\,mol\\
\text{ Lập tỉ lệ ta có }:\dfrac{{{n_{Al}}}}{2} > \dfrac{{{n_{{H_2}S{O_4}}}}}{3}(\dfrac{{0,3}}{2} > \dfrac{{0,1}}{3})\\
\Rightarrow \text{ Al dư }\\
{n_{Al}} \text{ dư }= 0,3 - \dfrac{{0,1 \times 2}}{3} = \dfrac{7}{{30}}\,mol\\
{m_{Al}} \text{ dư }= \dfrac{7}{{30}} \times 27 = 6,3g\\
b)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{{H_2}S{O_4}}}}}{3} = \dfrac{{0,1}}{3} = \dfrac{1}{{30}}\,mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{{30}} \times 342 = 11,4g\\
c)\\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,1\,mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l
\end{array}\)
