Đáp án:
\( {m_{Zn}} = 12,84{\text{ gam;}}{{\text{m}}_{Fe}} = 44,8{\text{ gam}}\)
\( {V_{{H_2}}} = 31,3{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(ZnO + {H_2}\xrightarrow{{{t^o}}}Zn + {H_2}O\)
\(F{e_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2Fe + 3{H_2}O\)
Ta có:
\({m_{ZnO}} = 80.20\% = 16{\text{ gam;}}{{\text{m}}_{F{e_2}{O_3}}} = 80 - 16 = 64{\text{ gam}}\)
\( \to {n_{ZnO}} = \frac{{16}}{{81}};{n_{F{e_2}{O_3}}} = \frac{{64}}{{56.2 + 16.3}} = 0,4{\text{ mol}}\)
\( \to {n_{Zn}} = {n_{ZnO}} = \frac{{16}}{{81}};{n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,8{\text{ mol}}\)
\( \to {m_{Zn}} = \frac{{16}}{{81}}.65 = 12,84{\text{ gam;}}{{\text{m}}_{Fe}} = 0,8.56 = 44,8{\text{ gam}}\)
\({n_{{H_2}}} = {n_{ZnO}} + 3{n_{F{e_2}{O_3}}} = \frac{{16}}{{81}} + 0,4.3 = \frac{{566}}{{405}}{\text{ mol}}\)
\( \to {V_{{H_2}}} = \frac{{566}}{{405}}.22,4 = 31,3{\text{ lít}}\)
