Bài 3:
a.
m ct HCl=$\frac{200.73}{100}$=146 g
n HCl=$\frac{146}{36,5}$=4 mol
CaCO3+2HCl→CaCl2+CO2↑+H2O
2 ← 4 → 2 2 mol
b.
-dd sau pứ:CaCl2
mdd sau=m CaCO3+mdd HCl-m CO2
=2.100+200-2.44=312 g
m ct CaCl2=2.111=222 g
⇒C% CaCl2=$\frac{222}{312}$.100≈71,15 %
---------------------Nguyễn Hoạt-------------------------