Em tham khảo nha :
\(\begin{array}{l}
1)\\
a)\\
{\rm{[}}O{H^ - }{\rm{]}} = {C_{{M_{KOH}}}} = 0,4M\\
pOH = - \log (0,4) = 0,4\\
pH = 14 - 0,4 = 13,6\\
b)\\
{\rm{[}}{H^ + }{\rm{]}} = 2{C_{{M_{{H_2}S{O_4}}}}} = 0,1M\\
pH = - \log (0,1) = 1\\
2)\\
{n_{C{O_2}}} = \dfrac{{8,8}}{{44}} = 0,2mol\\
{n_C} = {n_{C{O_2}}} = 0,2mol\\
{n_{{H_2}O}} = \dfrac{{3,6}}{{18}} = 0,2mol\\
{n_H} = 2{n_{{H_2}O}} = 0,4mol\\
BT\,Oxi:\\
{m_O} = 6 - 0,2 \times 12 - 0,4 \times 1 = 3,2\\
{n_O} = \dfrac{{3,2}}{{16}} = 0,2mol\\
CTHH:{C_x}{H_y}{O_z}\\
x:y:z = 0,2:0,4:0,1 = 1:2:1\\
CTDGN:C{H_2}O\\
3)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
hh:Al(a\,mol),Zn(b\,mol)\\
\left\{ \begin{array}{l}
27a + 65b = 11,9\\
\dfrac{3}{2}a + b = 0,4
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,1\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
{m_{Zn}} = 11,9 - 5,4 = 6,5g
\end{array}\)