Đáp án:

a)2,266l

b)0,083M

c) mAl2(SO4)3=65,195g

VH2=12,693l

#gaxinne

Giải thích các bước giải:

ở dưới

Em tham khảo nha:

\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{10,2}}{{27}} = \dfrac{{17}}{{45}}\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{17}}{{45}} \times \dfrac{3}{2} = \dfrac{{17}}{{30}}\,mol\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{\dfrac{{17}}{{30}}}}{{0,25}} = \dfrac{{34}}{{15}} \approx 2,267l\\
b)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{17}}{{45}} \times \dfrac{1}{2} = \dfrac{{17}}{{90}}\,mol\\
{C_M}A{l_2}{(S{O_4})_3} = \dfrac{{\dfrac{{17}}{{90}}}}{{\dfrac{{34}}{{15}}}} = 0,0833M\\
c)\\
{m_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{17}}{{90}} \times 342 = 64,6g\\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = \dfrac{{17}}{{30}}\,mol\\
{V_{{H_2}}} = \dfrac{{17}}{{30}} \times 22,4 = 12,693l
\end{array}\)