Em tham khảo nha:

\(\begin{array}{l}
a)\\
HCl:A\,xit\,clohidric\\
{n_{HCl}} = \dfrac{{33,6}}{{22,4}} = 1,5\,mol\\
{C_M}HCl = \dfrac{{1,5}}{3} = 0,5M\\
{C_\% }HCl = \dfrac{{1,5 \times 36,5}}{{1,5 \times 36,5 + 3000}} \times 100\%  = 1,79\% \\
b)\\
2K + 2{H_2}O \to 2KOH + {H_2}\\
KOH:Kali\,hidroxit\\
{n_K} = \dfrac{{15,6}}{{39}} = 0,4\,mol\\
{n_{{H_2}}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
{m_{{\rm{dd}}KOH}} = 15,6 + 250 - 0,2 \times 2 = 265,2g\\
{n_{KOH}} = {n_K} = 0,4\,mol\\
{C_M}KOH = \dfrac{{0,4}}{{0,25}} = 1,6M\\
{C_\% }KOH = \dfrac{{0,4 \times 56}}{{265,2}} \times 100\%  = 8,45\% \\
c)\\
{P_2}{O_5} + 3{H_2}O \to 2{H_3}P{O_4}:\\
{H_3}P{O_4}:Axit\,photphoric\\
{n_{{P_2}{O_5}}} = \dfrac{{14,2}}{{142}} = 0,1\,mol\\
{n_{{H_3}P{O_4}}} = 0,1 \times 2 = 0,2\,mol\\
{C_M}{H_3}P{O_4} = \dfrac{{0,2}}{{0,3}} = 0,67M\\
{m_{{\rm{dd}}{H_3}P{O_4}}} = 14,2 + 300 \times 1 = 314,2g\\
{C_\% }{H_3}P{O_4} = \dfrac{{0,2 \times 98}}{{314,2}} \times 100\%  = 6,24\% \\
d)\\
{n_{BaO}} = \dfrac{{13,7}}{{153}} = 0,09\,mol\\
{n_{Ba{{(OH)}_2}}} = {n_{BaO}} = 0,09\,mol\\
{C_M}Ba{(OH)_2} = \dfrac{{0,09}}{{0,5}} = 0,18M\\
{C_\% }Ba{(OH)_2} = \dfrac{{0,09 \times 171}}{{13,7 + 500}} \times 100\%  = 3\% 
\end{array}\)