Đáp án:

a) \({{\text{V}}_{{O_2}{\text{ dư}}}} = 26,88{\text{ lit}}\)

b) \({{\text{m}}_{CaC{O_3}}} = 15{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2KCl{O_3}\xrightarrow{{Mn{O_2},{t^o}}}2KCl + 3{O_2}\)

Ta có: 

\({n_{KCl{O_3}}} = \frac{{122,5}}{{39 + 35,5 + 16.3}} = 1{\text{ mol}} \to {{\text{n}}_{{O_2}}} = \frac{3}{2}{n_{KCl{O_3}}} = 1,5{\text{ mol}}\)

Đốt cháy metan

\(C{H_4} + 2{O_2}\xrightarrow{{}}C{O_2} + 2{H_2}O\)

Ta có: \({n_{C{H_4}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol < }}\frac{1}{2}{n_{{O_2}}}\)

\(\to {n_{{O_2}{\text{ dư}}}} = 1,5 - 2{n_{C{H_4}}} = 1,5 - 0,15.2 = 1,2{\text{ mol}} \to {{\text{V}}_{{O_2}{\text{ dư}}}} = 1,2.22,4 = 26,88{\text{ lít}}\)

Ta có: \({n_{C{O_2}}} = {n_{C{H_4}}} = 0,15{\text{ mol}}\)

\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)

\({n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,15{\text{ mol}} \to {{\text{m}}_{CaC{O_3}}} = 0,15.100 = 15{\text{ gam}}\).