\(CaCO_3+2HCl-->CaCl_2+CO_2+H_2O\left(1\right)\)
0,25_______0,5_________0,25_____0,25
\(n_{CaCO_3}=0,25\left(mol\right)\)
=> \(V_{CO_2}=0,25.22,4=5,6\left(l\right)\)
\(m_{d^2HCl}=\frac{0,5.36,5.100}{14,6}=125\left(g\right)\)
\(m_{d^2sau}=25+125-0,25.44=139\left(g\right)\)
=> \(C\%_{CaCl_2}=\frac{0,25.111}{139}.100=19,96\%\)