Đáp án:
a) ${m_{{K_2}O}} = 11,092g;{m_{CaO}} = 11,308g$
b) $C{\% _{KCl}} = 5,13\% ;C{\% _{CaC{l_2}}} = 6,55\% $
Giải thích các bước giải:
a)
${K_2}O + 2HCl \to 2KCl + {H_2}O$
$CaO + 2HCl \to CaC{l_2} + {H_2}O$
${n_{HCl}} = \dfrac{{320.7,3\% }}{{36,5}} = 0,64mol$
Gọi x, y là số mol ${K_2}O;CaO$
Ta có hpt: $\left\{ \begin{gathered} 2x + 2y = 0,64 \hfill \\ 94x + 56y = 22,4 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} x = 0,118 \hfill \\ y = 0,202 \hfill \\ \end{gathered} \right.$
$\begin{gathered} \Rightarrow {m_{{K_2}O}} = 0,118.94 = 11,092g \hfill \\ \Rightarrow {m_{CaO}} = 22,4 - 11,092 = 11,308g \hfill \\ \end{gathered} $
b)
Bảo toàn khối lượng: ${m_{ddsau}} = {m_{hh}} + {m_{ddHCl}} = 22,4 + 320 = 342,4g$
${n_{KCl}} = 2{n_{{K_2}O}} = 0,236mol;{n_{CaC{l_2}}} = {n_{CaO}} = 0,202mol$
$\begin{gathered} \Rightarrow C{\% _{KCl}} = \dfrac{{0,236.74,5}}{{342,4}}.100\% = 5,13\% \hfill \\ C{\% _{CaC{l_2}}} = \dfrac{{0,202.111}}{{342,4}}.100\% = 6,55\% \hfill \\ \end{gathered} $
