Đáp án:
\({m_{ZnO}}= 16,2{\text{ gam}}\)
\({{\text{V}}_{{O_2}}} = 2,24{\text{ lít}})
\({V_{C{O_2}}} = 1,12{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Zn + {O_2}\xrightarrow{{{t^o}}}2ZnO\)
Ta có:
\({n_{Zn}} = \frac{{13}}{{65}} = 0,2{\text{ mol = }}{{\text{n}}_{ZnO}}\)
\( \to {m_{ZnO}} = 0,2.(65 + 16) = 16,2{\text{ gam}}\)
\({n_{{O_2}}} = \frac{1}{2}{n_{Zn}} = 0,1{\text{ mol}} \to {{\text{V}}_{{O_2}}} = 0,1.22,4 = 2,24{\text{ lít}})
Đốt cháy \(CH_4\)
\(C{H_4} + 2{O_2}\xrightarrow{{{t^o}}}C{O_2} + 2{H_2}O\)
Ta có:
\({V_{C{H_4}}} > \frac{1}{2}{V_{{O_2}}}\)
Vậy \(CH_4\) dư.
\( \to {V_{C{O_2}}} = \frac{1}{2}{V_{{O_2}}} = 1,12{\text{ lít}}\)
