Đáp án:
V=3,36 lít
\({\text{C}}{{\text{\% }}_{Na}} = 12\% \)
Giải thích các bước giải:
\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)
Ta có:
\({n_{Na}} = \frac{{6,9}}{{23}} = 0,3{\text{ mol}} \to {{\text{n}}_{{H_2}}} = \frac{1}{2}{n_{Na}} = 0,15{\text{ mol}} \to {\text{V = 0}}{\text{,15}}{\text{.22}}{\text{,4 = 3}}{\text{,36 lít}}\)
BTKL: \({m_{dd{\text{ A}}}} = 6,9 + 93,4 - 0,15.2 = 100{\text{ gam}}\)
\({n_{NaOH}} = {n_{Na}} = 0,3{\text{ mol}} \to {{\text{m}}_{Na}} = 0,3.40 = 12{\text{ gam}} \to {\text{C}}{{\text{\% }}_{Na}} = \frac{{12}}{{100}} = 12\% \)