Đáp án:
\(\begin{array}{l} C\%_{CaCl_2}=19,96\%\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} PTHH:\\ CaO+2HCl\to CaCl_2+H_2O\ (1)\\ CaCO_3+2HCl\to CaCl_2+CO_2↑+H_2O\ (2)\\ n_{CO_2}=\dfrac{2,24}{22,4}=0,1\ mol.\\ Theo\ pt\ (2):\ n_{CaCO_3}=n_{CO_2}=0,1\ mol.\\ \Rightarrow m_{CaCO_3}=0,1\times 100=10\ g.\\ \Rightarrow m_{CaO}=15,6-10=5,6\ g.\\ \Rightarrow n_{CaO}=\dfrac{5,6}{56}=0,1\ mol.\\ \Rightarrow \sum n_{HCl}=2n_{CaCO_3}+2n_{CaO}=0,2+0,2=0,4\ mol.\\ \Rightarrow m_{\text{dd HCl}}=\dfrac{0,4\times 36,5}{14,6\%}=100\ g.\\ Rightarrow m_{\text{dd spư}}=m_{A}+m_{\text{dd HCl}}-m_{CO_2}\\ \Rightarrow m_{\text{dd spư}}=15,6+100-0,1\times 44=111,2\ g.\\ \sum n_{CaCl_2}=n_{CaO}+n_{CaCO_3}=0,1+0,1=0,2\ mol.\\ \Rightarrow C\%_{CaCl_2}=\dfrac{0,2\times 111}{111,2}\times 100\%=19,96\%\end{array}\)
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