Đáp án:
KOH
Giải thích các bước giải:
Gọi hidroxit có dạng \(M{(OH)_n}\)
Giả sử có 1 mol hidroxit.
\(M{(OH)_n} + nHCl\xrightarrow{{}}MC{l_n} + n{H_2}O\)
\(MC{l_n} + nAgN{O_3}\xrightarrow{{}}M{(N{O_3})_n} + nAgCl\)
Ta có:
\({n_{HCl}} = n{n_M} = n{\text{ mol}} \to {{\text{m}}_{HCl}} = 36,5n \to {m_{dd{\text{ HCl}}}} = \frac{{36,5n}}{{10\% }} = 365n{\text{ gam}}\)
\({n_{AgN{O_3}}} = n.{n_{MC{l_n}}} = n.{n_M} = n{\text{ mol}} \to {{\text{m}}_{AgN{O_3}}} = 170n{\text{ gam}} \to {{\text{m}}_{dd{\text{AgN}}{{\text{O}}_3}}} = \frac{{170n}}{{20\% }} = 850n{\text{ gam}}\)
\({n_{AgCl}} = {n_{HCl}} = n{\text{ mol}} \to {{\text{m}}_{AgCl}} = 143,5n{\text{ gam}}\)
BTKL:
\({m_{dd}} = {m_{MO{H_n}}} + {m_{dd{\text{HCl}}}} + {m_{dd\;{\text{AgN}}{{\text{O}}_3}}} - {m_{AgCl}} = 1.(M + 17n) + 365n + 850n - 143,5n = M + 1088,5n{\text{ gam}}\)
\({n_{M{{(N{O_3})}_n}}} = {n_M} = 1{\text{ mol}} \to {{\text{m}}_{M{{(N{O_3})}_n}}} = 1.(M + 62.n) = M + 62n\)
\( \to C{\% _{M{{(N{O_3})}_n}}} = \frac{{M + 62n}}{{M + 1088,5n}} = 8,965\% \to M = 39n \to n = 1;M = 39 \to K\)
Hidroxit là KOH
