Đáp án:
\({m_{N{a_2}C{O_3}}} = 7,42{\text{ gam}}\)
Giải thích các bước giải:
Ta có:
\({n_{C{O_2}}} = \frac{{1,568}}{{22,4}} = 0,07{\text{ mol}}\)
\({n_{NaOH}} = \frac{{6,4}}{{40}} = 0,16{\text{ mol}}\)
\( \to \frac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \frac{{0,16}}{{0,07}} > 2\)
Do vậy NaOH dư.
\(2NaOH + S{O_2}\xrightarrow{{}}N{a_2}S{O_3} + {H_2}O\)
\( \to {n_{NaOH{\text{ dư}}}} = 0,16 - 0,07.2 = 0,02{\text{ mol}}\)
\( \to {m_{NaOH{\text{ dư}}}} = 0,02.40 = 0,8{\text{ gam}}\)
\({n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,07{\text{ mol}}\)
\( \to {m_{N{a_2}C{O_3}}} = 0,07.(23.2 + 60) = 7,42{\text{ gam}}\)
