32)
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{HCl}} = 0,15.4 = 0,6{\text{ mol}} \to {{\text{n}}_{Zn}} = {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,3{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,3.22,4 = 6,72{\text{ lít;m = 0}}{\text{,3}}{\text{.65 = 19}}{\text{,5 gam}}\)
\({n_{ZnC{l_2}}} = {n_{Zn}} = 0,3{\text{ mol}} \to {{\text{C}}_{M\;{\text{ZnC}}{{\text{l}}_2}}} = \frac{{0,3}}{{0,15}} = 2M\)
Cho lượng HCl trên tác dụng với Al
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
Ta có:
\({n_{Al}} = \frac{{40,5}}{{27}} = 1,5{\text{ mol > }}\frac{1}{3}{n_{HCl}} \to {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,3{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)
33)
\(Mg + {H_2}S{O_4}\xrightarrow{{}}MgS{O_4} + {H_2}\)
\({n_{{H_2}}} = {n_{{H_2}S{O_4}}} = {n_{Mg}} = {n_{MgS{O_4}}} = \frac{{7,2}}{{24}} = 0,3{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)
\({V_{{H_2}S{O_4}}} = \frac{{0,3}}{2} = 0,15{\text{ lít = 150ml}}\)
Thể tích của dung dịch trước và sau phản ứng không đổi
\( \to {C_{M{\text{ Mg S}}{{\text{O}}_4}}} = \frac{{0,3}}{{0,15}} = 2M\)
\(MgS{O_4} + 2NaOH\xrightarrow{{}}Mg{(OH)_2} + N{a_2}S{O_4}\)
Ta có:
\({n_{NaOH}} = 0,2.2 = 0,4{\text{ mol < 2}}{{\text{n}}_{MgS{O_4}}} \to {n_{Mg{{(OH)}_2}}} = \frac{1}{2}{n_{NaOH}} = 0,2{\text{ mol}} \to {{\text{m}}_{Mg{{(OH)}_2}}} = 0,2.(24 + 17.2) = 11,6{\text{ gam}}\)
