Đáp án:
\({V_{C{O_2}{\text{ dư}}}} = 0,896{\text{ lít}}\)
\( {m_{N{a_2}C{O_3}}} = 14,84{\text{ gam}}\)
Giải thích các bước giải:
Ta có:
\({n_{C{O_2}}} = \frac{{3,136}}{{22,4}} = 0,14{\text{ mol}}\)
\({n_{NaOH}} = \frac{{12,8}}{{40}} = 0,32{\text{ mol}}\)
\( \to \frac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \frac{{0,32}}{{0,14}} > 2 \to NaOH\) dư.
\(2NaOH + C{O_2}\xrightarrow{{}}N{a_2}C{O_3} + {H_2}O\)
\( \to {n_{C{O_2}{\text{ dư}}}} = 0,32 - 0,14.2 = 0,04{\text{ mol}}\)
\( \to {V_{C{O_2}{\text{ dư}}}} =0,04.22,4 = 0,896{\text{ lít}}\)
\({n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,14{\text{ mol}}\)
\( \to {m_{N{a_2}C{O_3}}} = 0,14.(23.2 + 12 + 16.3) = 14,84{\text{ gam}}\)
