\(m_{HCl}=\dfrac{200.7,5\%}{100\%}=15\left(g\right)\)
\(n_{HCl}=\dfrac{15}{36,5}=0,41\left(mol\right)\)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
0,205 0,41 0,205 0,205 0,205 (mol)
\(m_{CaCO_3}=0,205.100=20,5\left(g\right)\)
\(m_{CaCl_2}=0,205.111=22,755\left(g\right)\)
\(m_{CO_2}=0,205.44=9,02\left(g\right)\)
\(m_{dd}=m_{CaCO_3}+m_{ddHCl}-m_{CO_2}=20,5+200-9,02=211,48\left(g\right)\)
\(C\%_{CaCl_2}=\dfrac{22,755.100}{211,48}=10,76\%\)