CaCO3(s)+2HCl(aq)>CaCl2(aq)+H2O(l)+CO2(g)
I will estimate all of these numbers; you need to go through and recalculate each.
mols CaCO3 = grams/molar mass = approx 0.30
mols HCl = grams/molar mass = approx 0.36
Convert mols CaCO3 to mols CaCl2 using the coefficients in the balanced equation.
CaCO3 to mols CaCl2 = approx 0.30
HCl to mols CaCl2 = approx 0.18
In limiting reagent (LR) problems the correct value is ALWAYS the smaller value and the reagent producing that value is the LR. So HCl is the LR and CaCO3 is in excess.
mols CaCl2 = 0.18 x molar mass CaCl2 = ?
How much of the CaCO3 is left unreacted? Using the coeffients again, convert mols HCl used to mols CaCO3. That's 0.36 mols HCl x (1 mol CaCO3/2 mols HCl) = 0.18 mols CaCO3. You started with approx 0.30 and used approx 0.18 so you have approx 0.12 mols CaCO3 remaining unreacted. Converted to grams that is g = mols x molar mass = ?
CaCO3(s)+2HCl(aq)>CaCl2(aq)+H2O(l)+CO2(g)
how many grams of calcium chloride will be produced when 30 g of calcium carbonate are combined with 13 g of hydrochloric acid?
Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?
2 answers
CaCO3+2HCI CaCL2+CO2+H2O