Đáp án:
\({V_{{O_2}}} = 7,28{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(4P + 5{O_2}\xrightarrow{{{t^o}}}2{P_2}{O_5}\)
Ta có:
\({n_P} = \frac{{6,2}}{{31}} = 0,2{\text{ mol}} \to {{\text{n}}_{{O_2}{\text{phản ứng}}}} = \frac{5}{4}{n_P} = 0,25{\text{ mol}} \to {{\text{n}}_{{O_2}{\text{ tham gia}}}} = 0,25.130\% = 0,325{\text{ mol}}\)
\( \to {V_{{O_2}}} = 0,325.22,4 = 7,28{\text{ lít}}\)
Ta có:
\({n_{{P_2}{O_5}}} = \frac{1}{2}{n_P} = 0,1{\text{ mol}} \to {{\text{m}}_{{P_2}{O_5}}} = 0,1.(31.2 + 16.5) = 14,2{\text{ gam}}\)
\({P_2}{O_5} + 23{H_2}O\xrightarrow{{}}2{H_3}P{O_4}\)
\({n_{{H_3}P{O_4}}} = 2{n_{{P_2}{O_5}}} = 0,2{\text{ mol}} \to {m_{{H_3}P{O_4}}} = 0,2.(3 + 31 + 16.4) = 19,6{\text{ gam}}\)
\({m_{dd{\text{ axit}}}} = 14,2 + 235,8 = 250{\text{ gam}}\)
\( \to C{\% _{{H_3}P{O_4}}} = \frac{{19,6}}{{250}} = 11,84\% \)
\({V_{dd}} = \frac{{250}}{{1,25}} = 200{\text{ ml = 0}}{\text{,2 lít}} \to {{\text{C}}_{M\;{{\text{H}}_3}P{O_4}}} = \frac{{0,2}}{{0,2}} = 1M\)
