Đáp án:
\({{\text{V}}_{{H_2}}} = 1,12{\text{ lít}}\)
\({\text{C}}{{\text{\% }}_{NaOH}} = 2\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)
Ta có:
\({n_{Na}} = \frac{{2,3}}{{23}} = 0,1{\text{ mol}} \to {{\text{n}}_{{H_2}}} = \frac{1}{2}{n_{Na}} = 0,05{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,05.22,4 = 1,12{\text{ lít}}\)
\({n_{NaOH}} = {n_{Na}} = 0,1{\text{ mol}} \to {{\text{m}}_{NaOH}} = 0,1.40 = 4{\text{ gam}} \to {\text{C}}{{\text{\% }}_{NaOH}} = \frac{4}{{200}} = 2\% \)