Đáp án:
a) Zn dư 6,5 gam
b) \({{\text{V}}_{{H_2}}} = 4,48{\text{ lit; }}{{\text{m}}_{ZnC{l_2}}} = 27,2{\text{ gam}}\)
c) \({{\text{m}}_{Fe}} = 8,4{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{Zn}} = \frac{{19,5}}{{65}} = 0,3{\text{ mol; }}{{\text{n}}_{HCl}} = \frac{{14,6}}{{36,5}} = 0,4{\text{ mol}}\)
Vì \({n_{HCl}} < 2{n_{Zn}}\) do vậy Zn dư
\(\to {n_{Zn{\text{ dư}}}} = 0,3 - \frac{1}{2}{n_{HCl}} = 0,3 - 0,2 = 0,1{\text{ mol}} \to {{\text{m}}_{Zn{\text{ dư}}}} = 0,1.65 = 6,5{\text{ gam}}\)
Ta có:
\({n_{{H_2}}} = {n_{ZnC{l_2}}} = \frac{1}{2}{n_{HCl}} = 0,2{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,2.22,4 = 4,48{\text{ lít; }}{{\text{m}}_{ZnC{l_2}}} = 0,2.(65 + 35,5.2) = 27,2{\text{ gam}}\)
\(F{e_3}{O_4} + 4{H_2}\xrightarrow{{}}3Fe + 4{H_2}O\)
\(\to {n_{Fe}} = \frac{3}{4}{n_{{H_2}}} = 0,15{\text{ mol}} \to {{\text{m}}_{Fe}} = 0,15.56 = 8,4{\text{ gam}}\)
