Đáp án:
\( {V_{{O_2}}} =7{\text{ lít}}\)
\({m_{{P_2}{O_5}}} =17,75{\text{ gam}}\)
\( {m_{{H_3}P{O_4}}} =24,5{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(4P + 5{{\text{O}}_2}\xrightarrow{{{t^o}}}2{P_2}{O_5}\)
Ta có:
\({n_P} = \dfrac{{7,75}}{{31}} = 0,25{\text{ mol}}\)
\( \to {n_{{O_2}}} = \dfrac{5}{4}{n_P} = 0,3125{\text{ mol}}\)
\( \to {V_{{O_2}}} = 0,3125.22,4 = 7{\text{ lít}}\)
Ta có:
\({n_{{P_2}{O_5}}} = \dfrac{1}{2}{n_P} = 0,125{\text{ mol}}\)
\( \to {m_{{P_2}{O_5}}} = 0,125.(31.2 + 16.5) = 17,75{\text{ gam}}\)
\(3{H_2}O + {P_2}{O_5}\xrightarrow{{}}2{H_3}P{O_4}\)
\( \to {n_{{H_3}P{O_4}}} = 2{n_{{P_2}{{\text{O}}_5}}} = 0,125.2 = 0,25{\text{ mol}}\)
\( \to {m_{{H_3}P{O_4}}} = 0,25.(3 + 31 + 16.4) = 24,5{\text{ gam}}\)