1.
$n_{HCl}=0,02.0,5=0,01(mol)$
$V_{dd}=100ml=0,1l$
$\to C_{M_{HCl}}=\dfrac{0,01}{0,1}=0,1M$
2.
$NaF$ không tạo kết tủa.
$n_{NaCl}=0,05.0,1=0,005(mol)$
$\to n_{AgCl\downarrow}=n_{NaCl}=0,005(mol)$
$\to m_{\downarrow}=0,005.143,5=0,7175g$
3.
$n_{CO_2}=\dfrac{0,448}{22,4}=0,02(mol)$
Bảo toàn $C$: $n_{CaCO_3}=n_{CO_2}=0,02(mol)$
$n_{CaCl_2}=\dfrac{3,33}{111}=0,03(mol)$
Bảo toàn $Ca$:
$n_{CaCl_2}=n_{CaO}+n_{CaCO_3}$
$\to n_{CaO}=n_{CaCl_2}-n_{CaCO_3}=0,01(mol)$
Hỗn hợp ban đầu có:
$m_{CaO}=0,01.56=0,56g$
$m_{CaCO_3}=0,02.100=2g$