Đáp án:
\({V_{{O_2}}} = 2,8{\text{ lit; }}{{\text{m}}_{{P_2}{O_5}}} = 7,1{\text{ gam; }}{{\text{m}}_{KMn{O_4}}} = 46,47{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(4P + 5{O_2}\xrightarrow{{}}2{P_2}{O_5}\)
Ta có: \({n_P} = \frac{{3,1}}{{31}} = 0,1{\text{ mol}} \to {{\text{n}}_{{O_2}}} = \frac{5}{4}{n_P} = 0,125{\text{ mol}} \to {{\text{V}}_{{O_2}}} = 0,125.22,4 = 2,8{\text{ lít}}\)
\({n_{{P_2}{O_5}}} = \frac{1}{2}{n_P} = 0,05{\text{ mol}} \to {{\text{m}}_{{P_2}{O_5}}} = 0,05.(31.2 + 16.5) = 7,1{\text{ gam}}\)
\(2KMn{O_4}\xrightarrow{{}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)
Vì quá trình thu khí hao thụt 15%
\(\to {n_{{O_2}{\text{ tạo ra}}}} = \frac{{0,125}}{{85\% }} = \frac{5}{{34}}{\text{ mol}} \to {{\text{n}}_{KMn{O_4}}} = 2{n_{{O_2}}} = \frac{5}{{17}}{\text{ mol}} \to {{\text{m}}_{KMn{O_4}}} = \frac{5}{{17}}.(39 + 55 + 16.4) = 46,47{\text{ gam}}\)
