Đáp án:
\(\begin{array}{l}
\% {m_{NaHC{O_3}}} = 53,33\% \\
\% {m_{{K_2}C{O_3}}} = 32,86\% \\
\% {m_{Mg{{(OH)}_2}}} = 13,81\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
NaHC{O_3} + HCl \to NaCl + C{O_2} + {H_2}O\\
{K_2}C{O_3} + 2HCl \to 2KCl + C{O_2} + {H_2}O\\
Mg{(OH)_2} + 2HCl \to MgC{l_2} + 2{H_2}O\\
{n_{HCl}} = 0,2 \times 1 = 0,2\,mol\\
{n_{C{O_2}}} = \dfrac{{2,464}}{{22,4}} = 0,11\,mol\\
hh:NaHC{O_3}(a\,mol),{K_2}C{O_3}(b\,mol),Mg{(OH)_2}(c\,mol)\\
84a + 138b + 58c = 12,6\\
a + 2b + 2c = 0,2\\
a + b = 0,11\\
\Rightarrow a = 0,08;b = c = 0,03\\
\% {m_{NaHC{O_3}}} = \dfrac{{0,08 \times 84}}{{12,6}} \times 100\% = 53,33\% \\
\% {m_{{K_2}C{O_3}}} = \dfrac{{0,03 \times 138}}{{12,6}} \times 100\% = 32,86\% \\
\% {m_{Mg{{(OH)}_2}}} = 100 - 53,33 - 32,86 = 13,81\%
\end{array}\)
