Đáp án:

Bạn tham khảo lời giải ở dưới nhé !

Giải thích các bước giải:

1.

\(\begin{array}{l}
Ba{(OH)_2} + 2HCl \to BaC{l_2} + 2{H_2}O\\
{n_{HCl}} = 0,2mol\\
a.\\
{n_{Ba{{(OH)}_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,1mol\\
 \to C{M_{Ba{{(OH)}_2}}} = \dfrac{{0,1}}{{0,15}} = 0,67M
\end{array}\)

 \(\begin{array}{l}
b.\\
{n_{BaC{l_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,1mol\\
 \to C{M_{BaC{l_2}}} = \dfrac{{0,1}}{{0,15 + 0,2}} = 0,29M\\
 \to {m_{BaC{l_2}}} = 20,8g
\end{array}\)

\(\begin{array}{l}
c.\\
BaC{l_2} + 2AgN{O_3} \to Ba{(N{O_3})_2} + 2AgCl\\
{n_{AgCl}} = 2{n_{BaC{l_2}}} = 0,2mol\\
 \to {m_{AgCl}} = 28,7g
\end{array}\)

2.

\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O
\end{array}\)

\(\begin{array}{l}
a.\\
{n_{Cu}} = 0,08mol\\
{n_{{H_2}}} = 0,1mol\\
 \to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,067mol\\
 \to {m_{Al}} = 1,809g\\
 \to {m_{F{e_2}{O_3}}} = 15 - (1,809 + 5) = 8,2g\\
 \to {n_{F{e_2}{O_3}}} = 0,05mol
\end{array}\)

\(\begin{array}{l}
b.\\
{n_{HCl}} = 3{n_{Al}} + 6{n_{F{e_2}{O_3}}} = 0,5mol\\
 \to C{M_{HCl}} = \dfrac{{0,5}}{{0,2}} = 2,5M
\end{array}\)

\(\begin{array}{l}
c.\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,067mol\\
{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,1mol\\
 \to C{M_{AlC{l_3}}} = \dfrac{{0,067}}{{0,2}} = 0,335M\\
 \to C{M_{FeC{l_3}}} = \dfrac{{0,1}}{{0,2}} = 0,5M
\end{array}\)