8)
Gọi:
\({n_{A{l_2}{O_3}}} = {n_{ZnO}} = x \to {n_{CuO}} = 1,5x\)
\( \to 102x + 81x + 80.1,5x = 30,3 \to x = 0,1{\text{ mol}}\)
Dùng hidro khử hỗn hợp oxit này
\(ZnO + {H_2}\xrightarrow{{{t^o}}}Zn + {H_2}O\)
\(CuO + {H_2}\xrightarrow{{{t^o}}}Cu + {H_2}O\)
\( \to {n_{{H_2}}} = {n_{ZnO}} + {n_{CuO}} = 0,1 + 0,15 = 0,25{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,25.22,4 = 5,6{\text{ lít}}\)
9)
Các phản ứng xảy ra:
\(ZnO + {H_2}\xrightarrow{{{t^o}}}Zn + {H_2}O\)
\(CuO + {H_2}\xrightarrow{{{t^o}}}Cu + {H_2}O\)
\(F{e_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2Fe + 3{H_2}O\)
Ta có:
\({n_{{H_2}O}} = \frac{{6,3}}{{18}} = 0,35{\text{ mol = }}{{\text{n}}_{{H_2}}}\)
BTKL:
\({m_X} + {m_{{H_2}}} = {m_Y} + {m_{{H_2}O}} \to m + 0,35.2 = 26,5 + 6,3 \to m = 32,1{\text{ gam}}\)
