$n_{CaCO_3}=\dfrac{10}{100}=0,1mol \\m_{HCl}=5\%.200=10g \\⇒n_{HCl}=\dfrac{10}{36,5}≈0,27mol \\PTHH :$
$CaCO_3 + 2HCl\to CaCl_2+H_2O+CO_2$
Theo pt : 1 mol 2 mol
Theo dbai : 0,1 mol 0,27 mol
Tỉ lệ : $\dfrac{0,1}{1}<\dfrac{0,27}{2}$
⇒Sau pư HCl dư
$Theo\ pt : \\n_{HCl\ pư}=2.n_{CaCO_3}=2.0,1=0,2mol \\⇒n_{HCl\ dư}=0,27-0,2=0,07mol \\⇒m_{HCl\ dư}=0,07.36,5=2,555g \\n_{CaCl_2}=n_{CO_2}=n_{CaCO_3}=0,1mol \\⇒m_{CaCl_2}=0,1.111=11,1g \\m_{dd\ spư}=10+200-0,1.44=205,6g \\⇒C\%_{HCl\ dư}=\dfrac{2,555}{205,6}.100\%=1,24\% \\C\%_{CaCl_2}=\dfrac{11,1}{205,6}.100\%=5,4\%$
