Đáp án:
\({m_{ZnC{l_2}}} = 27,2{\text{ gam;}}{{\text{V}}_{{H_2}}} = 4,48{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{Zn}} = \frac{{13}}{{65}} = 0,2{\text{ mol = }}{{\text{n}}_{ZnC{L_2}}} = {n_{{H_2}}}\)
\( \to {m_{ZnC{l_2}}} = 0,2.(65 + 35,5.2) = 27,2{\text{ gam;}}{{\text{V}}_{{H_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
\(CuO + {H_2}\xrightarrow{{}}Cu + {H_2}O\)
Ta có:
\({n_{CuO}} = \frac{{12}}{{64 + 16}} = 0,15{\text{ mol < }}{{\text{n}}_{{H_2}}} \to {H_2}\)
\( \to {n_{{H_2}{\text{ dư}}}} = 0,2 - 0,15 = 0,05{\text{ mol}} \to {{\text{m}}_{{H_2}{\text{ dư}}}} = 0,05.2 = 0,1{\text{ gam}}\)