Đáp án:

 1) mKClO3=43,365 gam

2) mMnO2=8,7 gam

Giải thích các bước giải:

 1) \(2KCl{O_3}\xrightarrow{{}}2KCl + 3{O_2}\)

Ta có:  \({n_{{O_2}}} = \frac{{14,4}}{{16.2}} = 0,45{\text{ mol}} \to {{\text{n}}_{KCl{O_3}}} = \frac{2}{3}{n_{{O_2}}} = 0,3{\text{ mol}} \to {{\text{n}}_{KCl{O_3}{\text{ thực tế cần}}}} = 0,3.118\%  = 0,354{\text{ mol}}\)

\(\to {m_{KCl{O_3}{\text{ can}}}} = 0,354.(39 + 35,5 + 16.3) = 43,365{\text{ gam}}\)

2)  \(2KMn{O_4}\xrightarrow{{}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)

\({n_{{O_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}}\)

Ta có: \({n_{Mn{O_2}}} = {n_{{O_2}}} = 0,1 \to {m_{Mn{O_2}}} = 0,1(55 + 16.2) = 8,7{\text{ gam}}\)