Đáp án:
\({{\text{m}}_{Zn}} = 6,5{\text{ gam;}}{{\text{V}}_{{H_2}}} = 2,24{\text{ lít}}\)
\({\text{C}}{{\text{\% }}_{ZnC{l_2}}} = 11,7\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{HCl}} = 0,1.2 = 0,2{\text{ mol}} \to {{\text{n}}_{Zn}} = {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,1{\text{ mol}} \to {{\text{m}}_{Zn}} = 0,1.65 = 6,5{\text{ gam;}}{{\text{V}}_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
\({m_{dd{\text{ HCl}}}} = 1,1.100 = 110{\text{ gam}} \to {{\text{m}}_{dd{\text{ sau phản ứng}}}} = 0,1.65 + 110 - 0,1.2 = 116,3{\text{ gam}}\)
\({n_{ZnC{l_2}}} = 0,1{\text{ mol}} \to {{\text{m}}_{ZnC{l_2}}} = 0,1.(65 + 35,5.2) = 13,6{\text{ gam}} \to {\text{C}}{{\text{\% }}_{ZnC{l_2}}} = \frac{{13,6}}{{116,3}} = 11,7\% \)