Đáp án:
\({m_{ZnC{l_2}}} = 8,575{\text{ gam}}\)
\({V_{{H_2}}} = 1,12{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{Zn}} = \frac{{6,5}}{{65}} = 0,1{\text{ mol}}\)
\({n_{HCl}} = \frac{{3,65}}{{36,5}} = 0,1{\text{ mol < 2}}{{\text{n}}_{Zn}}\)
Vậy \(Zn\) dư.
\( \to {n_{ZnC{l_2}}} = {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,05{\text{ mol}}\)
\( \to {m_{ZnC{l_2}}} = 0,05.(65 + 35,5.3) = 8,575{\text{ gam}}\)
\({V_{{H_2}}} = 0,05.22,4 = 1,12{\text{ lít}}\)
