1.Cho 6,5 gam Zn vao binh dung dung dich chua 0,25 mol HCl thu duoc ZnCl2 va giai phong hi hidro a.chat nao con du sau phan ung?du bao nhieu mol b.ti

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.

\( \to {n_{F{e_3}{O_4}}} = \frac{1}{3}{n_{Fe}} = 0,1{\text{ mol}} \to {{\text{m}}_{F{e_3}{O_4}}} = 0,1.(56.3 + 16.4) = 23,2{\text{ gam}}\)