Đáp án:
\(\begin{array}{l}
{m_{{K_2}C{O_3}}} = 62,1g\\
{m_{KHC{O_3}}} = 30g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{n_{C{O_2}}} = \dfrac{{16,8}}{{22,4}} = 0,75mol\\
{n_{KOH}} = 0,6 \times 2 = 1,2mol\\
T = \dfrac{{{n_{KOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{1,2}}{{0,75}} = 1,6\\
\Rightarrow {K_2}C{O_3},KHC{O_3}\\
2KOH + C{O_2} \to {K_2}C{O_3} + {H_2}O\\
KOH + C{O_2} \to KHC{O_3}\\
hh:{K_2}C{O_3}(a\,mol),KHC{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
2a + b = 1,2\\
a + b = 0,75
\end{array} \right.\\
\Rightarrow a = 0,45;b = 0,3\\
{m_{{K_2}C{O_3}}} = 0,45 \times 138 = 62,1g\\
{m_{KHC{O_3}}} = 0,3 \times 100 = 30g
\end{array}\)
