Formula: CaCO3(s) + HCl(aq) --> CO2(g) + CaCl2(aq) + H2O

Question

How many grams of CO2 can be made by reacting 125 g of CaCO3 with 125 grams of HCl?

Answer 1

CaCO3(s) + 2HCl(aq) --> CO2(g) + CaCl2(aq) + H2O
Note that I balanced the equation.
This is a limiting reagent (LR) problem. You know that when amounts are given for BOTH reactants instead of one reactant and an excess of the other one. So this essentially is two stoichiometry problems in one. I do these the long way but there is a shorter way.
mols CaCO3 = g/molar mass = 125/100 = 1.25
mols HCl = 125/36.5 = 3.42
Now convert each of these into how many mols CO2 would be formed IF the other reagent was in excess.
For CaCO3 that will be 1.25 x (1 mol CO2/1 mol CaCO3) = 1.25 mols CO2.
For HCl that will be 3.42 x (2 mols HCl/1 mol CO2) = 3.42 x 2 = 6.84 mols CO2.
In LR problems the SMALLER number is always the correct one since that one will be the product formed and the other reagent will react as needed and the remainder will be in excess and be left over.
So 1.25 mol CO2 will be formed.
grams CO2 = mols CO2 x molar mass CO2 = 1.25 x 44 = ?

Answer 2

Where did you get 125?

Answer 3

Nevermind my last question