Đáp án:

\(\begin{array}{l}
22,?\\
23,A\\
24,B\\
25,B
\end{array}\)

Giải thích các bước giải:

22, Bạn xem lại đề dùm mình nhé!!!

23,

\(\begin{array}{l}
{H_2}S{O_4} + Ba{(OH)_2} \to BaS{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{{100 \times 24,5}}{{100 \times 98}} = 0,25mol\\
{n_{Ba{{(OH)}_2}}} = \dfrac{{50 \times 17,1}}{{100 \times 171}} = 0,05mol\\
 \to {n_{{H_2}S{O_4}}} > {n_{Ba{{(OH)}_2}}}
\end{array}\)

Suy ra \({H_2}S{O_4}\) dư 

\(\begin{array}{l}
 \to {n_{{H_2}S{O_4}}}dư= 0,25 - 0,05 = 0,2mol\\
 \to {n_{BaS{O_4}}} = {n_{Ba{{(OH)}_2}}} = 0,05mol\\
{m_{dd}} = 100 + 50 - 0,05 \times 233 = 138,35g\\
 \to C{\% _{{H_2}S{O_4}}}dư= \dfrac{{0,2 \times 98}}{{138,35}} \times 100\%  = 14,2\% 
\end{array}\)

24, \({N_2} + {O_2} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} 2NO\)

25,

\(\begin{array}{l}
N{H_4}Cl + NaOH \to NaCl + N{H_3} + {H_2}O\\
{n_{N{H_4}Cl}} = 0,15mol\\
 \to {n_{N{H_3}}} = {n_{N{H_4}Cl}} = 0,15mol\\
 \to {V_{N{H_3}}} = 3,36l
\end{array}\)