Đáp án:

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Giải thích các bước giải:

\(\begin{array}{l}
5.\\
Ba{(OH)_2} + 2HCl \to BaC{l_2} + 2{H_2}O\\
{m_{HCl}} = \dfrac{{150 \times 14,6\% }}{{100\% }} = 21,9g\\
 \to {n_{HCl}} = 0,6mol\\
 \to {n_{BaC{l_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,3mol\\
 \to a = {V_{BaC{l_2}}} = \frac{{0,3}}{1} = 0,3l\\
6.\\
AgN{O_3} + HCl \to AgCl + HN{O_3}\\
{n_{HCl}} = 0,04mol\\
 \to {n_{AgCl}} = {n_{HCl}} = 0,04mol\\
 \to {m_{AgCl}} = 5,74g\\
7.\\
KOH + HCl \to KCl + {H_2}O\\
{n_{HCl}} = 0,6mol\\
 \to {n_{KOH}} = {n_{HCl}} = 0,6mol\\
 \to a = C{M_{KOH}} = \dfrac{{0,6}}{{0,2}} = 3M\\
8.\\
KOH + HCl \to KCl + {H_2}O\\
{m_{KOH}} = \dfrac{{11,2\%  \times 200}}{{100\% }} = 22,4g\\
 \to {n_{KOH}} = 0,4mol\\
 \to {n_{KCl}} = {n_{KOH}} = 0,4mol\\
 \to {m_{KCl}} = 29,8g
\end{array}\)

 \(\begin{array}{l}
9.\\
M + C{l_2} \to MC{l_2}\\
{n_{C{l_2}}} = 0,2mol\\
 \to {n_M} = {n_{C{l_2}}} = \dfrac{{13}}{M} = 0,2\\
 \to M = 65(Zn)\\
10.\\
M + C{l_2} \to MC{l_2}\\
{n_{C{l_2}}} = 0,3mol\\
 \to {n_M} = {n_{C{l_2}}} = \dfrac{{7,2}}{M} = 0,3\\
 \to M = 24(Mg)
\end{array}\)