a)
Ta có: \({n_{AlC{l_3}}} = 0,2.1 = 0,2{\text{ mol;}}{{\text{n}}_{NaOH}} = 0,1.3,6 = 0,36{\text{ mol}}\)
\( \to \frac{{{n_{NaOH}}}}{{{n_{AlC{l_3}}}}} = \frac{{0,36}}{{0,2}} = 1,8 < 3\)
Vậy phản ứng xảy ra:
\(AlC{l_3} + 3NaOH\xrightarrow{{}}Al{(OH)_3} + 3NaCl\)
\( \to {n_{Al{{(OH)}_3}}} = \frac{1}{3}{n_{NaOH}} = 0,12{\text{ mol}} \to {m_{Al{{(OH)}_3}}} = 0,12.78 = 9,36{\text{ gam}}\)
b)
Ta có: \({n_{AlC{l_3}}} = 0,2.1 = 0,2{\text{ mol;}}{{\text{n}}_{NaOH}} = 0,2.3,4 = 0,68{\text{ mol}} \to \frac{{{n_{NaOH}}}}{{{n_{AlC{l_3}}}}} > 3\)
Phản ứng xảy ra:
\(AlC{l_3} + 3NaOH\xrightarrow{{}}Al{(OH)_3} + 3NaCl\)
\(Al{(OH)_3} + NaOH\xrightarrow{{}}NaAl{O_2} + 2{H_2}O\)
\( \to {n_{NaOH{\text{ hòa tan kết tủa}}}} = {n_{NaAl{O_2}}} = {n_{NaOH}} - 3{n_{AlC{l_3}}} = 0,68 - 0,2.3 = 0,08{\text{ mol}}\)
\( \to {n_{Al{{(OH)}_3}}} = {n_{AlC{l_3}}} - {n_{NaAl{O_2}}} = 0,2 - 0,08 = 0,12{\text{ mol}} \to {{\text{m}}_{Al{{(OH)}_3}}} = 0,12.78 = 9,36{\text{ gam}}\)
c)
Ta có:
\({n_{ZnC{l_2}}} = 0,1.1 = 0,1{\text{ mol;}}{{\text{n}}_{KOH}} = 0,24.1 = 0,24{\text{ mol}} \to \frac{{{n_{KOH}}}}{{{n_{ZnC{l_2}}}}} > 2\)
Vậy phản ứng xảy ra:
\(ZnC{l_2} + 2KOH\xrightarrow{{}}Zn{(OH)_2} + 2KCl\)
\(Zn{(OH)_2} + 2KOH\xrightarrow{{}}{K_2}Zn{O_2} + 2{H_2}O\)
\( \to {n_{KOH{\text{ hòa tan kết tủa}}}} = 0,24 - 0,1.2 = 0,04{\text{ mol}} \to {{\text{n}}_{{K_2}Zn{O_2}}} = \frac{1}{2}{n_{KOH{\text{ hòa tan kết tủa}}}} = 0,02{\text{ mol}}\)
\( \to {n_{Zn{{(OH)}_2}}} = {n_{ZnC{l_2}}} - {n_{{K_2}Zn{O_2}}} = 0,1 - 0,02 = 0,08{\text{ mol}} \to {{\text{m}}_{Zn{{(OH)}_2}}} = 0,08.(65 + 17.2) = 7,92{\text{ gam}}\)
