Đáp án:

\(\begin{array}{l}
15\,A\\
16\,C\\
17\,C\\
18\,A\\
19\,D\\
20\,C\\
21\,C\\
22\,C\\
23\,B\\
24\,C\\
25\,B\\
26\,C\\
27\,A\\
28\,D\\
29\,B
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
25)\\
2{H_2} + {O_2} \to 2{H_2}O\\
\dfrac{{{V_{{H_2}}}}}{2} < \dfrac{{{V_{{O_2}}}}}{1} \Rightarrow  \text{ $O_2$ dư }\\
26)\\
{m_{{\rm{dd}}}} = 30 + 170 = 200g\\
{C_\% }NaCl = \dfrac{{30}}{{200}} \times 100\%  = 15\% \\
27)\\
{m_{{H_2}O}} = 876 \times 1 = 876g\\
{m_{{\rm{dd}}}} = 124 + 876 = 1000g\\
{n_{N{a_2}O}} = \dfrac{{124}}{{62}} = 2\,mol\\
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{NaOH}} = 2{n_{N{a_2}O}} = 4\,mol\\
{C_\% }NaOH = \dfrac{{4 \times 40}}{{1000}} \times 100\%  = 16\% \\
28)\\
{m_{{\rm{dd}}}} = 1 \times 98 + 18 = 116g\\
{C_\% }{H_2}S{O_4} = \dfrac{{98}}{{116}} \times 100\%  = 84,48\% \\
29)\\
S{O_3} + {H_2}O \to {H_2}S{O_4}\\
{V_{{\rm{dd}}}} = {V_{{H_2}O}} = \dfrac{{100}}{1} = 100ml = 0,1l\\
{n_{{H_2}S{O_4}}} = {n_{S{O_3}}} = \dfrac{{12}}{{80}} = 0,15\,mol\\
{C_M}{H_2}S{O_4} = \dfrac{{0,15}}{{0,1}} = 1,5M
\end{array}\)

Đáp án:

 15A

16C

17C

18A

19D

20C

21C

22C

23B

24D

25B

26B

27A phải là 876gam nước nha

28D

29B Phải là 100ml nước nha