Đáp án:
\( V= 2,24{\text{ lít}}\)
\({V_{NaOH}} {\text{ = 600 ml}}\)
\({m_{BaS{O_4}}} = 93,2{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(F{\text{e}} + {H_2}S{O_4}\xrightarrow{{}}F{\text{e}}S{O_4} + {H_2}\)
Ta có:
\({n_{F{\text{e}}}} = \frac{{5,6}}{{56}} = 0,1{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = 0,2.2 = 0,4{\text{ mol > }}{{\text{n}}_{F{\text{e}}}}\)
Vậy \(H_2SO_4\) dư
\( \to {n_{{H_2}}} = {n_{F{\text{e}}}} = 0,1{\text{ mol}} \to V={{\text{V}}_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
Trung hoà axit dư
\({H_2}S{O_4} + 2NaOH\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
\({n_{{H_2}S{O_4}{\text{ dư}}}} = 0,4 - 0,1 = 0,3{\text{ mol}}\)
\( \to {n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,3.2 = 0,6{\text{ mol}}\)
\( \to {V_{NaOH}} = \frac{{0,6}}{1} = 0,6{\text{ lít = 600 ml}}\)
\(F{\text{e}}S{O_4} + BaC{l_2}\xrightarrow{{}}BaS{O_4} + F{\text{e}}C{l_2}\)
\({H_2}S{O_4} + BaC{l_2}\xrightarrow{{}}BaS{O_4} + 2HCl\)
\( \to {n_{BaS{O_4}}} = {n_{F{\text{e}}S{O_4}}} + {n_{{H_2}S{O_4}}} = 0,4{\text{ mol}}\)
\( \to {m_{BaS{O_4}}} = 0,4.233 = 93,2{\text{ gam}}\)
