Đáp án:
\( C{\% _{NaHC{O_3}}} = 8,046\% \)
Giải thích các bước giải:
Ta có:
\({n_{C{O_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol}}\)
\({m_{NaOH}} = 200.4\% = 8{\text{ gam}}\)
\( \to {n_{NaOH}} = \frac{8}{{40}} = 0,2{\text{ mol}}\)
\( \to \frac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \frac{{0,2}}{{0,2}} = 1\) nên phản ứng tạo muối \(NaHCO_3\)
\(NaOH + C{O_2}\xrightarrow{{}}NaHC{O_3}\)
Ta có:
\({m_{NaHC{O_3}}} = {n_{C{O_2}}} = 0,2{\text{ mol}}\)
\( \to {m_{NaHC{O_3}}} = 0,2.(23 + 1 + 12 + 16.3) = 16,8{\text{ gam}}\)
BTKL:
\({m_{dd}} = {m_{C{O_2}}} + {m_{dd{\text{ NaOH}}}} = 0,2.44 + 200 = 208,8{\text{ gam}}\)
\( \to C{\% _{NaHC{O_3}}} = \frac{{16,8}}{{208,8}} = 8,046\% \)
