$n_{CaCO_3}=\dfrac{10}{100}=0,1mol \\m_{HCl}=114,1.8\%=9,128g \\⇒n_{HCl}=\dfrac{9,128}{36,5}=0,25mol \\PTHH :$
$CaCO_3 + 2HCl\to CaCl_2+H_2O+CO_2$
Theo pt : 1 mol 2 mol
Theo đbài : 0,1 mol 0,25 mol
Tỉ lệ : $\dfrac{0,1}{1}<\dfrac{0,25}{2}$
⇒Sau phản ứng HCl còn dư
a.Theo pt :
$n_{CO_2}=n_{CaCO_3}=0,1mol$
$⇒m_{CO_2}=0,1.44=4,4g$
b.Theo pt :
$n_{HCl\ pư}=2.n_{CaCO_3}=2.0,1=0,2mol \\⇒n_{HCl\ dư}=0,25-0,2=0,05mol \\⇒m_{HCl\ dư}=0,05.36,5=1,825g \\n_{CaCl_2}=n_{CaCO_3}=0,1mol \\⇒m_{CaCl_2}=0,1.111=11,1g \\m_{dd\ spư}=10+114,1-4,4=119,7g \\⇒C\%_{CaCl_2}=\dfrac{11,1}{119,7}.100\%=9,27\% \\C\%_{HCl\ dư}=\dfrac{1,825}{119,7}.100\%=1,52\%$
