Đáp án:
\( C{\% _{KOH}} = 5,3\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2K + 2{H_2}O\xrightarrow{{}}2KOH + {H_2}\)
Ta có:
\({n_K} = \frac{{7,8}}{{39}} = 0,2{\text{ mol = }}{{\text{n}}_{KOH}}\)
\({n_{{H_2}}} = \frac{1}{2}{n_K} = 0,1{\text{ mol}}\)
BTKL:
\({m_K} + {m_{{H_2}O}} = {m_{dd}} + {m_{{H_2}}}\)
\( \to 7,8 + 203,6 = {m_{dd}} + 0,1.2 \to {m_{dd}} = 211,2{\text{ gam}}\)
Ta có:
\({m_{KOH}} = 0,2.(39 + 17) = 11,2{\text{ gam}}\)
\( \to C{\% _{KOH}} = \frac{{{m_{KOH}}}}{{{m_{dd}}}} = \frac{{11,2}}{{211,2}} = 5,3\% \)