Đáp án:
\({{\text{V}}_{{H_2}}} = 2,24{\text{ lít}}\)
\( {m_{Zn{{(OH)}_2}}}= 9,9{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)
Ta có:
\({n_{Na}} = \frac{{4,6}}{{23}} = 0,2{\text{ mol = }}{{\text{n}}_{NaOH}}\)
\({n_{{H_2}}} = \frac{1}{2}{n_{Na}} = 0,1{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
\({n_{Zn{{(N{O_3})}_2}}} = 0,2.0,5 = 0,1{\text{ mol}}\)
\( \to \frac{{{n_{NaOH}}}}{{{n_{Zn{{(N{O_3})}_2}}}}} = \frac{{0,2}}{{0,1}} = 2\) nên phản ứng vừa đủ.
\(Zn{(N{O_3})_2} + 2NaOH\xrightarrow{{}}Zn{(OH)_2} + 2NaN{O_3}\)
\( \to {n_{Zn{{(OH)}_2}}} = {n_{Zn{{(N{O_3})}_2}}} = 0,1{\text{ mol}}\)
\( \to {m_{Zn{{(OH)}_2}}} = 0,1.(65 + 17.2) = 9,9{\text{ gam}}\)
