Đáp án:

\({V_{dd{\text{ BaC}}{{\text{l}}_2}}} = 0,47272{\text{ lít}}\)

Giải thích các bước giải:

 Ta có: 

\({n_{ZnS{O_4}}} = \frac{{40,24}}{{65 + 96}} = 0,25{\text{ mol}}\)

\( \to {C_{M{\text{ Zn S}}{{\text{O}}_4}}} = \frac{{0,25}}{{0,4}} = 0,625M\)

\( \to [Z{n^{2 + }}] = [S{O_4}^{2 - }] = {C_{M{\text{ ZnS}}{{\text{O}}_4}}} = 0,625M\)

Cho NaOH vào

\(ZnS{O_4} + 2NaOH\xrightarrow{{}}Zn{(OH)_2} + N{a_2}S{O_4}\)

\( \to {n_{NaOH}} = 2{n_{ZnS{O_4}}} = 0,25.2 = 0,5{\text{ mol}}\)

\( \to {V_{NaOH}} = \frac{{0,5}}{5} = 0,1{\text{ lít}}\)

\(BaC{l_2} + ZnS{O_4}\xrightarrow{{}}BaS{O_4} + ZnC{l_2}\)

\( \to {n_{BaC{l_2}}} = {n_{ZnS{O_4}}} = 0,25{\text{ mol}}\)

\( \to {m_{BaC{l_2}}} = 0,25.(137 + 35,5.2) = 52{\text{ gam}}\)

\( \to {m_{dd{\text{ BaC}}{{\text{l}}_2}}} = \frac{{52}}{{10\% }} = 520{\text{ gam}}\)

\( \to {V_{dd{\text{ BaC}}{{\text{l}}_2}}} = \frac{{520}}{{1,1}} = 472,72ml = 0,47272{\text{ lít}}\)