Gọi \(\left\{{}\begin{matrix}n_{Ca}:x\left(mol\right)\\n_{CaCO3}:y\left(mol\right)\end{matrix}\right.\)
Ta có:
\(n_{HCl}=0,2.2=0,4\left(mol\right)\)
PTHH:
\(Ca+2HCl\rightarrow CaCl_2+H_2\left(1\right)\)
x______2x______x_________
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\left(2\right)\)
y_______2y_________y_______________
Giải hệ PT:
\(\left\{{}\begin{matrix}40x+100y=14\\2x+2y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Ca}=\frac{0,1.40}{14}.100\%=28,57\%\\\%m_{CaCO3}=100\%-28,57\%=71,48\%\end{matrix}\right.\)
\(CM_{HCl}=2M\)
Theo PTHH 1: \(n_{CaCl2}=n_{Ca}=0,1\left(mol\right)\)
Theo PTHH 2: \(n_{CaCl2}=n_{CaCO3}=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{CaCl2}=0,1+0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{CaCl2}=0,2.111=22,2\left(g\right)\)
