Đáp án:
\({\text{V = 2}}{\text{,24 lít; }}{{\text{C}}_{M{\text{ NaOH}}}}{\text{ = 2M;C}}{{\text{\% }}_{NaOH}} = 7,66\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)
Ta có:
\({n_{Na}} = \frac{{4,6}}{{23}} = 0,2{\text{ mol}} \to {{\text{n}}_{{H_2}}} = \frac{1}{2}{n_{Na}} = 0,1{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
\({V_{{H_2}O}} = {V_{dd{\text{ A}}}} = \frac{{100}}{1} = 100{\text{ ml = 0}}{\text{,1 lit}}\)
\({n_{NaOH}} = {n_{Na}} = 0,2{\text{ mol}} \to {{\text{C}}_{M{\text{ NaOH}}}} = \frac{{0,2}}{{0,1}} = 2M\)
BTKL: \({m_{Na}} + {m_{{H_2}O}} = {m_{dd{\text{ A}}}} + {m_{{H_2}}} \to {m_{dd{\text{ A}}}} = 4,6 + 100 - 0,1.2 = 104,4{\text{ gam}}\)
\({m_{NaOH}} = 0,2.40 = 8{\text{ gam}} \to {\text{C}}{{\text{\% }}_{NaOH}} = \frac{8}{{104,4}} = 7,66\% \)
