Giải thích các bước giải:
2,
\(\begin{array}{l}
2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\
{n_{C{l_2}}} = \dfrac{{P \times V}}{{R \times T}} = \dfrac{{0,2 \times 3,075}}{{0,082 \times (273 + 27)}} = 0,025mol\\
\to {n_{KMn{O_4}}} = \dfrac{2}{5}{n_{C{l_2}}} = 0,01mol \to m = 1,58g\\
\to {n_{HCl}} = \dfrac{{16}}{5}{n_{C{l_2}}} = 0,08mol \to m = 2,92g
\end{array}\)
3,
\(\begin{array}{l}
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
{n_{C{l_2}}} = \dfrac{{P \times V}}{{R \times T}} = \dfrac{{0,5 \times 2,24}}{{0,082 \times 273}} = 0,05mol\\
{n_{Mn{O_2}}} = {n_{C{l_2}}} = 0,05mol \to m = 3,45g\\
{n_{HCl}} = 4{n_{C{l_2}}} = 0,2mol \to m = 7,3g
\end{array}\)
5,
\(\begin{array}{l}
2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\
2NaOH + C{l_2} \to NaCl + NaClO + {H_2}O\\
{n_{NaOH}} = 0,5mol \to {n_{C{l_2}}} = \frac{1}{2}{n_{NaOH}} = 0,25mol\\
\to {n_{KMn{O_4}}} = \frac{2}{5}{n_{C{l_2}}} = 0,1mol \to m = 15,8g
\end{array}\)
6,
\(\begin{array}{l}
2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
{n_{FeC{l_3}}} = 0,1mol \to {n_{C{l_2}}} = \dfrac{3}{2}{n_{FeC{l_3}}} = 0,15mol\\
{n_{HCl}} = \dfrac{{16}}{5}{n_{C{l_2}}} = 0,48mol \to {V_{HCl}} = \dfrac{n}{{CM}} = 480ml
\end{array}\)
7,
\(\begin{array}{l}
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
2NaI + C{l_2} \to 2NaCl + {I_2}\\
{n_{{I_2}}} = 0,05mol = {n_{C{l_2}}} \to {n_{HCl}} = 4{n_{C{l_2}}} = 0,2mol \to m = 7,3g
\end{array}\)
8,
\(\begin{array}{l}
2Al + 3C{l_2} \to 2AlC{l_3}\\
{n_{AlC{l_3}}} = 0,2mol \to {n_{C{l_2}}} = \dfrac{3}{2}{n_{AlC{l_3}}} = 0,3mol \to {V_{C{l_2}}} = 6,72l
\end{array}\)
