Đáp án đúng: B
600 ml.
Đặt nFe3O4 = a mol ; nCuO =b mol=> mX = 232a +80b =19,6 (1)Các phản ứngVới HCl$\displaystyle \begin{array}{l}\text{F}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ + 8HCl }\xrightarrow{{}}\text{ FeC}{{\text{l}}_{\text{2}}}\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ + 2FeC}{{\text{l}}_{\text{3}}}\text{ }\!\!~\!\!\text{ + 4}{{\text{H}}_{\text{2}}}\text{O}\\\,\,\,\,\text{a}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{8a}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{2a}\\\text{CuO }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ + 2HCl }\!\!~\!\!\text{ }\xrightarrow{{}}\text{ CuC}{{\text{l}}_{\text{2}}}\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ + }{{\text{H}}_{\text{2}}}\text{O}\\\,\,\text{b}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{2b}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{b}\end{array}$ Với H2S$\displaystyle \begin{array}{l}{{\text{H}}_{\text{2}}}\text{S }\!\!~\!\!\text{ + 2FeC}{{\text{l}}_{\text{3}}}\text{ }\!\!~\!\!\text{ }\xrightarrow{{}}\text{ 2FeC}{{\text{l}}_{\text{2}}}\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ S }\!\!~\!\!\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ }\!\!~\!\!\text{ 2 HCl}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{2a}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{a}\\{{\text{H}}_{\text{2}}}\text{S }\!\!~\!\!\text{ + CuC}{{\text{l}}_{\text{2}}}\text{ }\!\!~\!\!\text{ }\xrightarrow{{}}\text{ CuS }\!\!~\!\!\text{ }\!\!~\!\!\text{ + 2HCl}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{b}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{b}\end{array}$ Kết tủa là CuS và S => m kết tủa = 32a + 96b =11,2 (2)giải hệ pt (1) và (2) => a =0,05 mol và b =0,1 mol=> nHCl =8a +2b = 8.0,05+2.0,1 =0, 6 mol => VHCl = 0,6/1=0,6 lít = 600 ml
