\(n_{CaCO3}=\frac{12.6}{100}=0.126mol\)
Ta có: \(\%HCl=\frac{m_{ct}}{m_{dd}}\cdot100\)
\(\Leftrightarrow7.3=\frac{m_{ct}}{250}\cdot100\)
\(\Leftrightarrow m_{ct}=18,25\left(g\right)\)
\(\rightarrow n_{HCl}=\frac{18.25}{36.5}=0.5mol\)
PT: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
TBR: 1 mol : 2mol
TPT: 0.126 mol : 0.5mol
Lập tỉ lệ: \(\frac{0.126}{1}< \frac{0.5}{2}\)( HCl dư. tính toán dựa vào CaCO\(_3\))
\(\rightarrow n_{CaCO3}=n_{CaCl2}=0.126mol\)
\(\rightarrow m_{CaCl_2}=0.126\cdot111=13.986\left(g\right)\)
