Đáp án:
\( {m_{ZnC{l_2}}} = 34{\text{ gam}}\)
\({V_{{H_2}}} = 5,6{\text{ lít}}\)
\( {m_{CuO}} = 20{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{Zn}} = \frac{{16,25}}{{65}} = 0,25{\text{ }}mol = {n_{ZnC{l_2}}} = {n_{{H_2}}}\)
\( \to {m_{ZnC{l_2}}} = 0,25.(65 + 35,5.2) = 34{\text{ gam}}\)
\({V_{{H_2}}} = 0,25.22,4 = 5,6{\text{ lít}}\
Cho \(H_2\) khử \(CuO\)
\(CuO + {H_2}\xrightarrow{{{t^o}}}Cu + {H_2}O\)
Ta có:
\({n_{CuO}} = {n_{{H_2}}} = 0,25{\text{ mol}}\)
\( \to {m_{CuO}} = 0,25.(64 + 16) = 20{\text{ gam}}\)
