Đáp án:
m=4,7 gam
Giải thích các bước giải:
Ta có:
\({P_2}{O_5} + 3{H_2}O\xrightarrow{{}}2{H_3}P{O_4}\)
Ta có:
\({n_{{P_2}{O_5}}} = \frac{{2,13}}{{31.2 + 16.5}} = 0,015{\text{ mol}}\)
\( \to {n_{{H_3}P{O_4}}} = 2{n_{{P_2}{O_5}}} = 0,03{\text{ mol}}\)
\({n_{NaOH}} = 0,08.1 = 0,08{\text{ mol}} \to \frac{{{n_{NaOH}}}}{{{n_{{H_3}P{O_4}}}}} = \frac{{0,08}}{{0,03}} = 2,67\)
Phản ứng xảy ra:
\(3NaOH + {H_3}P{O_4}\xrightarrow{{}}N{a_3}P{O_4} + 3{H_2}O\)
\(2NaOH + {H_3}P{O_4}\xrightarrow{{}}N{a_2}HP{O_4} + 2{H_2}O\)
\( \to {n_{{H_2}O}} = {n_{NaOH}} = 0,08{\text{ mol}}\)
BTKL:
\({m_{{H_3}P{O_4}}} + {m_{NaOH}} = m + {m_{{H_2}O}}\)
\( \to 0,03.98 + 0,08.40 = m + 0,08.18 \to m = 4,7{\text{ gam}}\)
