Em tham khảo nha :
\(\begin{array}{l}
1)\\
a)\\
4Na + {O_2} \to 2N{a_2}O\\
b)\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
c)\\
A{l_2}{(S{O_4})_3} + 3BaC{l_2} \to 3BaS{O_4} + 2AlC{l_3}\\
d)\\
2Al{(OH)_3} \to A{l_2}{O_3} + 3{H_2}O\\
2)\\
a)\\
2Al{(OH)_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 6{H_2}O\\
b)\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
c)\\
3KOH + {H_3}P{O_4} \to {K_3}P{O_4} + 3{H_2}O\\
d)\\
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\
3)\\
CTHH:{C_x}{H_y}\\
M = 13{M_{{H_2}}} = 26dvC\\
\% C = \dfrac{{x{M_C}}}{M} \times 100\% = 92,3\% \\
\Rightarrow x = \dfrac{{26 \times 0,923}}{{12}} = 2\\
M = 2{M_C} + y{M_H} = 26\\
\Rightarrow y = 26 - 24 = 2\\
\Rightarrow CTHH:{C_2}{H_2}\\
4)\\
a)\\
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\
b)\\
{n_{{C_2}{H_4}}} = \dfrac{m}{M} = \dfrac{{2,8}}{{28}} = 0,1mol\\
{n_{{O_2}}} = 3{n_{{C_2}{H_4}}} = 0,3mol\\
{m_{{O_2}}} = n \times m = 0,3 \times 32 = 9,6g\\
5)\\
CTHH:N{a_x}{O_y}\\
M = 62dvC\\
\% O = \dfrac{{y{M_O}}}{M} \times 100\% = 25\% \\
\Rightarrow y = \dfrac{{0,25 \times 62}}{{16}} = 1\\
M = x{M_{Na}} + {M_O} = 62\\
\Rightarrow x = \dfrac{{62 - 16}}{{23}} = 2\\
\Rightarrow CTHH:N{a_2}O
\end{array}\)
