Đáp án:
1) a) H2 dư 0,05 mol
b) V H2=2,24 lít
2) mFe3O4=23,2 gam
Giải thích các bước giải:
1) \(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có: \({n_{Zn}} = \frac{{6,5}}{{65}} = 0,1{\text{ < }}\frac{1}{2}{{\text{n}}_{{\text{HCl}}}}\) nên HCl dư
\({n_{HCl{\text{ phản ứng}}}} = 2{n_{Zn}} = 0,1.2 = 0,2{\text{ mol}} \to {{\text{n}}_{HCl{\text{ dư}}}} = 0,25 - 0,2 = 0,05{\text{ mol }}\)
\({n_{{H_2}}} = {n_{Zn}} = 0,1{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
2) \(3Fe + 2{O_2}\xrightarrow{{}}F{e_3}{O_4}\)
\({n_{Fe}} = \frac{{16,8}}{{56}} = 0,3{\text{ mol; }}{{\text{n}}_{{O_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}} \to {{\text{n}}_{{O_2}}} > \frac{2}{3}{n_{Fe}}\) nên O2 dư.
\( \to {n_{F{e_3}{O_4}}} = \frac{1}{3}{n_{Fe}} = 0,1{\text{ mol}} \to {{\text{m}}_{F{e_3}{O_4}}} = 0,1.(56.3 + 16.4) = 23,2{\text{ gam}}\)