Phản ứng xảy ra:
\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
Ta có:
\(n_{MgCO3}=\frac{1,68}{24+60}=0,02\left(mol\right)\)
\(n_{Fe2O3}=\frac{1,6}{56.2+16.3}=0,01\left(mol\right)\)
Theo phản ứng:
\(n_{CO2}=n_{MgCO3}=0,02\left(mol\right);n_{HCl}=2n_{MgCO3}+6n_{Fe2O3}=0,02.2+0,01.6=0,1\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,1.36,5=3,65\left(g\right)\)
\(\Rightarrow m_{Dd_{HCl}}=\frac{3,65}{7,3\%}=50\left(g\right)\)
BTKL: \(m_{dd_{spu}}=1,68+1,6+50-0,02.44=52,4\left(g\right)\)
Dung dịch chứa MgCl2 0,02 mol và FeCl3 0,02 mol.
\(m_{MgCl2}=0,02.\left(24+35,5.2\right)=1,9\left(g\right)\Rightarrow C\%_{MgCl2}=\frac{1,9}{52,4}=3,625\%\)
\(m_{FeCl3}=0,02.\left(56+35,5.3\right)=3,25\left(g\right)\Rightarrow C\%_{FeCl3}=\frac{3,25}{52,4}=6,2\%\)
