Đáp án:
\({V_{C{O_2}}} = 4,48{\text{ lít}}\)
\({V_{kk}} = 44,8{\text{ lít}}\)
\({m_{N{a_2}C{O_3}}}= 5,3{\text{ gam}}\)
\({m_{NaHC{O_3}}} = 12,6{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(C{H_4} + 2{O_2}\xrightarrow{{{t^o}}}C{O_2} + 2{H_2}O\)
Ta có:
\({V_{C{O_2}}} = {V_{C{H_4}}} = 4,48{\text{ lít}}\)
\({V_{{O_2}}} = 2{V_{C{H_4}}} = 8,96{\text{ lít}}\)
\( \to {V_{kk}} = 5{V_{{O_2}}} = 8,96.5 = 44,8{\text{ lít}}\)
Ta có:
\({n_{C{O_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol;}}{{\text{n}}_{NaOH}} = 0,5.0,5 = 0,25{\text{ mol}}\)
\( \to \frac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \frac{{0,25}}{{0,2}} = 1,25\) nên tạo 2 muối.
\(2NaOH + C{O_2}\xrightarrow{{}}N{a_2}C{O_3} + {H_2}O\)
\(NaOH + C{O_2}\xrightarrow{{}}NaHC{O_3}\)
\( \to {n_{N{a_2}C{O_3}}} = {n_{NaOH}} - {n_{C{O_2}}} = 0,25 - 0,2 = 0,05{\text{ mol}}\)
\({n_{NaHC{O_3}}} = 0,2 - 0,05 = 0,15{\text{ mol}}\)
\( \to {m_{N{a_2}C{O_3}}} = 0,05.106 = 5,3{\text{ gam}}\)
\({m_{NaHC{O_3}}} = 0,15.84 = 12,6{\text{ gam}}\)
