Calculate the number of moles of calcium
chloride, CaCl2, that can be obtained from 25 g
of limestone, CaCO3, in the presence of excess
hydrogen chloride, HCl. (Ca = 40, C = 12,0 =
16, H = 1, Cl = 35.5)
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
mols CaCO3 = grams/molar mass = 25/100 = 0.25.
One look at the equation and you can see that 1 mol CaCO3 produces 1 mol CaCl2 so you should get 0.25 mols CaCl2 if the procedure is 100% efficient.
CaCO3(s)+ 2HCl(aq)> CaCl2(aq)+CO2(g)+H2O(l)
RFM of CaCO3= 40+12+(3×16)= 40+12+48= 100
no of moles of CaCO3= 25/100
=0.25 moles
from the equation, mole ratio of CaCO3 to CaCl2 is 1:1 therefore, the number of moles of calcium chloride= 0.25 moles