Đáp án:
\(x=0,7;y=1,1\)
Giải thích các bước giải:
TN1:
\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
\(2KOH + {H_2}S{O_4}\xrightarrow{{}}{K_2}S{O_4} + 2{H_2}O\)
Ta có:
\({n_{{H_2}S{O_4}}} = 0,03x{\text{ mol;}}{{\text{n}}_{NaOH}} = 0,02y{\text{ mol;}}{{\text{n}}_{KOH}} = 0,01.2 = 0,02{\text{ mol}}\)
\( \to 2{n_{{H_2}S{O_4}}} = {n_{NaOH}} + {n_{KOH}} \to 0,03x.2 = 0,02y + 0,02\)
TN2:
\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
\(NaOH + HCl\xrightarrow{{}}NaCl + {H_2}O\)
Ta có:
\({n_{NaOH}} = 0,03y;{n_{{H_2}S{O_4}}} = 0,02x;{n_{HCl}} = 0,005.1 = 0,005{\text{ mol}}\)
\( \to {n_{NaOH}} = 2{n_{{H_2}S{O_4}}} + {n_{HCl}} \to 0,03y = 0,02x.2 + 0,005\)
Giải được:
\(x=0,7;y=1,1\)
