Đáp án:
\(\% {m_{NaHC{O_3}}} = 50,9\% ;\% {m_{Ca{{(HC{O_3})}_2}}} = 49,1\% \)
Giải thích các bước giải:
Gọi số mol \(NaHC{O_3};{\text{ Ca(HC}}{{\text{O}}_3}{)_2}\) lần lượt là x, y.
Phản ứng xảy ra:
\(2NaHC{O_3}\xrightarrow{{}}N{a_2}C{O_3} + C{O_2} + {H_2}O\)
\(Ca{(HC{O_3})_2}\xrightarrow{{}}CaO + 2C{O_2} + {H_2}O\)
\(\to {n_{N{a_2}C{O_3}}} = \frac{1}{2}{n_{NaHC{O_3}}} = 0,5x;{\text{ }}{{\text{n}}_{CaO}} = {n_{Ca{{(HC{O_3})}_2}}} = y\)
\(\to 106.0,5x + 56y = 16,2\)
\(N{a_2}C{O_3} + 2HCl\xrightarrow{{}}2NaCl + C{O_2} + {H_2}O\)
\(CaO+ 2HCl\xrightarrow{{}}CaC{l_2} + {H_2}O\)
\(\to {n_{C{O_2}}} = {n_{N{a_2}C{O_3}}} = 0,5x = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}}\)
Vậy x=0,2; y=0,1.
\(\to {m_{NaHC{O_3}}} = 84x = 16,8{\text{ gam; }}{{\text{m}}_{Ca{{(HC{O_3})}_2}}} = 0,1.(40 + 61.2) = 16,2{\text{ gam}}\)
\(\to \% {m_{NaHC{O_3}}} = \frac{{16,8}}{{16,8 + 16,2}} = 50,9\% \to \% {m_{Ca{{(HC{O_3})}_2}}} = 49,1\% \)
