Đáp án:
a) V O2=7,4978 lít
b) mKClO3=27,336 gam
Giải thích các bước giải:
Phản ứng xảy ra
\(4Al + 3{O_2}\xrightarrow{{}}2A{l_2}{O_3}\)
Ta có: \({n_{Al}} = \frac{{12,05}}{{27}} \to {n_{{O_2}}} = \frac{3}{4}{n_{Al}} = \frac{{241}}{{720}}{\text{ mol}} \to {{\text{V}}_{{O_2}}} = \frac{{241}}{{720}}.22,4 = 7,4978{\text{lít}}\)
\(2KCl{O_3}\xrightarrow{{}}2KCl + 3{O_2}\)
\(\to {n_{KCl{O_3}}} = \frac{2}{3}{n_{{O_2}}} = \frac{{241}}{{1080}}{\text{ mol}} \to {{\text{m}}_{KCl{O_3}}} = \frac{{241}}{{1080}}.(39 + 35,5 + 16.3) = 27,336{\text{ gam}}\)