a, PTHH : \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
\(n_{CaCO_3}=\frac{m}{M}=\frac{10}{100}=0,1\left(mol\right)\)
\(n_{HCl}=C_M.V=2.0,02=0,04\left(mol\right)\)
- Theo PTHH : \(n_{CaCO_3}=\frac{1}{2}n_{HCl}=0,04.\frac{1}{2}=0,02\left(mol\right)\)
=> Sau phản ứng CaCO3 còn dư ( dư \(0,1-0,02=0,08\left(mol\right)\) ), HCl phản ứng hết .
- Theo PTHH : \(n_{CO_2}=\frac{1}{2}n_{HCl}=\frac{1}{2}.0,04=0,02\left(mol\right)\)
=> \(V_{CO_2}=n.22,4=0,02.22,4=0,448\left(l\right)\)
b, - Theo PTHH : \(n_{CaCl_2}=\frac{1}{2}n_{HCl}=\frac{1}{2}0,04=0,02\left(mol\right)\)
=> \(\left\{{}\begin{matrix}C_{MCaCl_2}=\frac{n}{V}=\frac{0,02}{0,02}=1\left(M\right)\\C_{MCaCO_3}=\frac{n}{V}=\frac{0,08}{0,02}=4\left(M\right)\end{matrix}\right.\)
