Đáp án:
\( C{\% _{NaOH}} = 3,9\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)
Ta có:
\({n_{Na}} = \frac{{4,6}}{{23}} = 0,2{\text{ mol = }}{{\text{n}}_{NaOH}}\)
\({n_{{H_2}}} = \frac{1}{2}{n_{Na}} = 0,1{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
\({m_{NaOH}} = 0,2.40 = 8{\text{ gam}}\)
BTKL:
\({m_{Na}} + {m_{{H_2}O}} = {m_{dd}} + {m_{{H_2}}}\)
\( \to 4,6 + 200 = {m_{dd}} + 0,1.2 \to {m_{dd}} = 204,4{\text{ gam}}\)
\( \to C{\% _{NaOH}} = \frac{8}{{204,4}} = 3,9\% \)