Đáp án:
\( {V_{{H_2}}} = 22,4{\text{ lít}}\)
\( C{\% _{NaOH}} = 7,66\% \)
\({D_{dd}} = 1,08{\text{ g/ml}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)
Ta có:
\({n_{Na}} = \frac{{46}}{{23}} = 2{\text{ mol}} \to {{\text{n}}_{{H_2}}} = \frac{1}{2}{n_{Na}} = 1{\text{ mol}}\)
\( \to {V_{{H_2}}} = 1.22,4 = 22,4{\text{ lít}}\)
Bảo toàn khối lượng:
\({m_{Na}} + {m_{{H_2}O}} = {m_{dd}} + {m_{{H_2}}}\)
\( \to 46 + 1000 = {m_{dd}} + 1.2 \to {m_{dd}} = 1044{\text{ gam}}\)
\({n_{NaOH}} = {n_{Na}} = 2{\text{ mol}} \to {{\text{m}}_{NaOH}} = 2.40 = 80{\text{ gam}}\)
\( \to C{\% _{NaOH}} = \frac{{80}}{{1044}}.100\% = 7,66\% \)
\({D_{dd}} = \frac{{{m_{dd}}}}{{{V_{dd}}}} = \frac{{1044}}{{966}} = 1,08{\text{ g/ml}}\)
