Đáp án:
\(\begin{array}{l} a,\ C_{M_{Ca(OH)_2}}=2,5\ M.\\ b,\ m_{CaCO_3}=25\ g.\\ c,\ m_{\text{dd HCl}}=91,25\ g.\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\ PTHH:CO_2+Ca(OH)_2\to CaCO_3↓+H_2O\\ n_{CO_2}=\dfrac{5,6}{22,4}=0,25\ mol.\\ Theo\ pt:\ n_{Ca(OH)_2}=n_{CO_2}=0,25\ mol.\\ ⇒C_{M_{Ca(OH)_2}}=\dfrac{0,25}{0,1}=2,5\ M.\\ b,\ Theo\ pt:\ n_{CaCO_3}=n_{CO_2}=0,25\ mol.\\ ⇒m_{CaCO_3}=0,25\times 100=25\ g.\\ c,\ PTHH:Ca(OH)_2+2HCl\to CaCl_2+2H_2O\\ Theo\ pt:\ n_{HCl}=2n_{Ca(OH)_2}=0,5\ mol.\\ ⇒m_{\text{dd HCl}}=\dfrac{0,5\times 36,5}{20\%}=91,25\ g.\end{array}\)
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