Em tham khảo nha :
\(\begin{array}{l}
3)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = \dfrac{{6,5}}{{65}} = 0,1mol\\
{m_{HCl}} = \dfrac{{200 \times 7,3}}{{100}} = 14,6g\\
{n_{HCl}} = \dfrac{{14,6}}{{36,5}} = 0,4mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,4}}{2} \Rightarrow HCl\text{ dư}\\
{n_{{H_2}}} = {n_{Zn}} = 0,1mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
4)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{Al}} = \dfrac{{10,8}}{{27}} = 0,4mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{300 \times 9,8}}{{100}} = 29,4g\\
{n_{{H_2}S{O_4}}} = \dfrac{{29,4}}{{98}} = 0,3mol\\
\dfrac{{0,4}}{2} > \dfrac{{0,3}}{3} \Rightarrow Al\text{ dư}\\
{n_{{H_2}}} = {n_{ {H_2}S{O_4}}} = 0,3mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
5)\\
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
b)\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
c)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,3mol\\
{m_{{H_2}S{O_4}}} = 0,3 \times 98 = 29,4g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{29,4 \times 100}}{{19,6}} = 150g\\
{V_{{H_2}S{O_4}}} = \dfrac{{150}}{{1,14}} = 131,58ml
\end{array}\)
