g CaCO3 reacted = 4.568 g - 2.678g = ?g CaCO3 used in the titration.
mols CaCO3 = gCaCO3 used/molar mass =?
mols HCl = twice that (from the balanced equation coefficients)
M HCl = mols HCl/L HCl.
Reaction:
CaCO3 + 2 HCl -> CaCl2 + CO2 + H2O
Calculate the molarity of hydrochloric acid solution if an initial mass of 4.568g of CaCO3 was reacted with 25.0mL of the acid and 2.678g of CaCO3 remain after the reaction is complete.
2 answers
1.890CaCO3/1 x 1mole CaCO3/100.09 x2molesHCl/1mole CaCO3 x1/0.025
=1.5
=1.5
1 answer