Đáp án:
\(\% {m_{N{a_2}C{O_3}}} = 42,4\% ; \% {m_{NaCl}} = 57,6\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(N{a_2}C{O_3} + 2HCl\xrightarrow{{}}2NaCl + C{O_2} + {H_2}O\)
\({n_{C{O_2}}} = \frac{{0,448}}{{22,4}} = 0,02{\text{ mol}}\)
\( \to {n_{HCl}} = 2{n_{C{O_2}}} = 0,04{\text{ mol}}\)
\( \to {C_{M{\text{ HCl}}}} = \frac{{0,04}}{{0,02}} = 2M\)
\({n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,02{\text{ mol}}\)
\( \to {m_{N{a_2}C{O_3}}} = 0,02.(23.2 + 12 + 16.3) = 2,12{\text{ gam}}\)
\( \to \% {m_{N{a_2}C{O_3}}} = \frac{{2,12}}{5} = 42,4\% \to \% {m_{NaCl}} = 57,6\% \)
