19C

20D

21B

22B

23D

24C

35B

26D

27B

28A

29C

30A

Em tham khảo nha :

\(\begin{array}{l}
19)\\
C\\
20)\\
D\\
S + {O_2} \to S{O_2}\\
21)\\
B\\
22)\\
B\\
CaO + {H_2}O \to Ca{(OH)_2}\\
23)\\
D\\
24)\\
C\\
25)\\
B\\
{C_{{M_{NaCl}}}} = \dfrac{2}{{1,25}} = 1,6M\\
26)\\
D\\
{M_X} = 1,375{M_{{O_2}}} = 44dvC\\
{M_{C{O_2}}} = 12 + 32 = 44dvC\\
27)\\
B\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
{n_{C{H_4}}} = \dfrac{{1,6}}{{16}} = 0,1mol\\
{n_{{O_2}}} = 2{n_{C{H_4}}} = 0,2mol\\
{m_{{O_2}}} = 0,2 \times 32 = 6,4g\\
28)\\
A\\
C + {O_2} \to C{O_2}\\
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_C} = \dfrac{{1,2}}{{12}} = 0,1mol\\
{n_{{O_2}}} = {n_C} = 0,1mol\\
{n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,2mol\\
{m_{KMn{O_4}}} = 0,2 \times 158 = 31,6g\\
29)\\
C\\
2M + {O_2} \to 2MO\\
{n_M} = \dfrac{{6,5}}{{{M_M}}}mol\\
{n_{MO}} = \dfrac{{8,1}}{{{M_M} + 16}}mol\\
{n_M} = {n_{MO}} \Rightarrow \dfrac{{6,5}}{{{M_M}}} = \dfrac{{8,1}}{{{M_M} + 16}}\\
 \Rightarrow {M_M} = 65dvC\\
 \Rightarrow M:Zn\\
30)\\
\\
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{N{a_2}O}} = \frac{{6,2}}{{62}} = 0,1mol\\
{n_{NaOH}} = 2{n_{N{a_2}O}} = 0,2mol\\
{m_{NaOH}} = 0,2 \times 40 = 8g\\
C\%  = \dfrac{8}{{6,2 + 43,8}} \times 100\%  = 16\% 
\end{array}\)