Đáp án:
\({m_{Zn}} = 1,3{\text{ gam}}; {{\text{m}}_{ZnO}} = 3,24{\text{ gam}}\)
\({C_{M{\text{ ZnC}}{{\text{l}}_2}}} = 0,6M;{C_{M{\text{ HCl}}}} = 0,3M\)
Giải thích các bước giải:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
\(ZnO + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}O\)
Ta có:
\({n_{{H_2}}} = \frac{{0,448}}{{22,4}} = 0,02{\text{mol = }}{{\text{n}}_{Zn}}\)
\( \to {m_{Zn}} = 0,02.65 = 1,3{\text{ gam}} \to {{\text{m}}_{ZnO}} = 4,54 - 1,3 = 3,24{\text{ gam}}\)
\( \to {n_{ZnO}} = \frac{{3,24}}{{65 + 16}} = 0,04{\text{ mol}}\)
\({n_{HCl}} = 0,1.1,5 = 0,15{\text{ mol > 2}}{{\text{n}}_{Zn}} + 2{n_{ZnO}}\)
\({n_{ZnC{l_2}}} = {n_{Zn}} + {n_{ZnO}} = 0,02 + 0,04 = 0,06{\text{ mol}}\)
\( \to {n_{HCl{\text{ dư}}}} = 0,15 - 0,02.2 - 0,04.2 = 0,03{\text{ mol}}\)
\( \to {C_{M{\text{ ZnC}}{{\text{l}}_2}}} = \frac{{0,06}}{{0,1}} = 0,6M;{C_{M{\text{ HCl}}}} = \frac{{0,03}}{{0,1}} = 0,3M\)
