Câu 1.
Ta có:
\({m_{NaOH}} = 44.10\% = 4,4{\text{ gam}} \\\to {{\text{n}}_{NaOH}} = \dfrac{{4,4}}{{40}} = 0,11{\text{ mol}}\)
Ta có:
\({m_{{H_3}P{O_4}}} = 10.39,2\% = 3,92{\text{ gam}} \\\to {{\text{n}}_{{H_3}P{O_4}}} = \dfrac{{3,92}}{{98}} = 0,04{\text{ mol}}\)
\( \to \dfrac{{{n_{KOH}}}}{{{n_{{H_3}P{O_4}}}}} = \dfrac{{0,11}}{{0,04}} = 2,75\)
Vậy phản ứng tạo 2 muối \(Na_3PO_4;Na_2HPO_4\)
\(3NaOH + {H_3}P{O_4}\xrightarrow{{}}N{a_3}P{O_4} + 3{H_2}O\)
\(2NaOH + {H_3}P{O_4}\xrightarrow{{}}N{a_2}HP{O_4} + 2{H_2}O\)
\( \to {n_{{H_2}O}} = {n_{NaOH}} = 0,11{\text{ mol}}\)
Bảo toàn khối lượng:
\({m_{NaOH}} + {m_{{H_3}P{O_4}}} = {m_{muối}} + {m_{{H_2}O}}\)
\( \to 4,4 + 3,92 = {m_{muối}} + 0,11.18 \to {m_{muối}} = 6,34{\text{ }}gam\)
Câu 2.
Ta có:
\({n_{N{H_3}}} = \dfrac{{6,72}}{{22,4}} = 0,3{\text{ mol;}}\\{{\text{n}}_{{H_3}P{O_4}}} = \dfrac{{9,8}}{{98}} = 0,1{\text{ mol}}\)
\( \to \dfrac{{{n_{N{H_3}}}}}{{{n_{{H_3}P{O_4}}}}} = \dfrac{{0,3}}{{0,1}} = 3\)
Phản ứng xảy ra:
\(3N{H_3} + {H_3}P{O_4}\xrightarrow{{}}{(N{H_4})_3}P{O_4}\)
\( \to {n_{{{(N{H_4})}_3}P{O_4}}} = {n_{{H_3}P{O_4}}} = 0,1{\text{ mol}}\)
\( \to {m_{{{(N{H_4})}_3}P{O_4}}} = 0,1.(18.3 + 95) = 14,9{\text{ gam}}\)
