\(n_{CaCO_3}=0,05\left(mol\right)\)
\(n_{HCl}=\dfrac{100.3,65\%}{36,5.100\%}=0,1\left(mol\right)\)
\(CaCO_3+2HCl-->CaCl_2+H_2O+CO_2\uparrow\)
\(\dfrac{0,05}{1}=\dfrac{0,1}{2}\) => 2 chất hết
dd sau phản ứng CaCl2
\(C\%CaCl_2=\dfrac{0,1.36,5}{5+100-0,05.44}.100\%\approx3,55\%\)