Đáp án:
\(C{\% _{KOH}} = 10\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2K + 2{H_2}O\xrightarrow{{}}2KOH + {H_2}\)
Ta có:
\({n_K} = \frac{{9,75}}{{39}} = 0,25{\text{ mol = }}{{\text{n}}_{KOH}}\)
\({n_{{H_2}}} = \frac{1}{2}{n_K} = 0,125{\text{ mol}}\)
\({m_{{H_2}O}} = 130,5.1 = 130,5{\text{ gam}}\)
BTKL:
\({m_K} + {m_{{H_2}O}} = {m_{dd}} + {m_{{H_2}}}\)
\( \to 9,75 + 130,5 = {m_{dd}} + 0,125.2 \to {m_{dd}} = 140{\text{ gam}}\)
\({m_{KOH}} = 0,25.(39 + 17) = 14{\text{ gam}}\)
\(C{\% _{KOH}} = \frac{{14}}{{140}} = 10\% \)