Giả sử \(m_{dd\left(HCl\right)}=45,625\left(g\right)\)
\(\Rightarrow n_{HCl}=\frac{45,625.80\%}{36,5}=1\left(mol\right)\)
\(FeCO_3+2HCl\rightarrow FeCl_2+CO_2+H_2O\)
\(\left\{{}\begin{matrix}n_{HCl\left(dư\right)}=1-2x\left(mol\right)\\n_{CO2}=x\left(mol\right)\end{matrix}\right.\)
\(m_{dd\left(spu\right)}=m_{FeCO3}+45,625-m_{CO2}=72x+45,625\)
\(\Rightarrow\frac{36,5.\left(1-2x\right)}{72x+45,625}=0,371\)
\(\Leftrightarrow x=0,2\left(mol\right)\)
\(\Rightarrow C\%_{FeCl2}=\frac{0,2.127.100}{72.0,2+45,625}=42,32\%\)
Y có 0,6 mol HCl và 0,2 mol FeCl2
\(m_{dd}=72.0,2+45,625=60,025\left(g\right)\)
\(BaCO_3+2HCl\rightarrow BaCl_2+CO_2+H_2O\)
Gọi y là mol BaCO3 thêm vào
\(n_{HCl}=0,6-2y\left(mol\right)\)
\(n_{CO2}=y\left(mol\right)\)
\(m_{dd\left(spu\right)}=m_{BaCO3}+60,025-m_{CO2}=153y+60,025\)
\(\Rightarrow\frac{36,5.\left(0,6-2y\right)}{153y+60,025}=0,197\)
\(\Leftrightarrow y=0,1\)
\(m_{dd\left(spu\right)}=153.0,1+60,025=75,325\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl2}=\frac{0,2.127.100}{75,325}=33,72\%\\C\%_{BaCl2}=\frac{0,1.208.100}{75,325}=27,61\%\end{matrix}\right.\)
