Đáp án:
\({m_{HCl{\text{ dư}}}} = 3,65{\text{ gam}}\)
\({V_{{H_2}}} = 1,12{\text{ lít}}\)
\({m_{ZnC{l_2}}} = 6,8{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{Zn}} = \frac{{3,25}}{{65}} = 0,05{\text{ mol;}}{{\text{n}}_{HCl}} = \frac{{7,3}}{{36,5}} = 0,2{\text{ mol}}\)
\( \to {n_{HCl}} > 2{n_{Zn}}\) nên \(HCl\) dư
\( \to {n_{HCl{\text{ dư}}}} = 0,2 - 0,05.2 = 0,1{\text{ mol}}\)
\( \to {m_{HCl{\text{ dư}}}} = 0,1.36,5 = 3,65{\text{ gam}}\)
\({n_{{H_2}}} = {n_{ZnC{l_2}}} = {n_{Zn}} = 0,05{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,05.22,4 = 1,12{\text{ lít}}\)
\({m_{ZnC{l_2}}} = 0,05.(65 + 35,5.2) = 6,8{\text{ gam}}\)
