Calculate how many grams of the product(s) would be produced by complete reaction of 0.125 mole of the first reactant.

CaCO3 + 2HCl --> CaCl2 + CO2 + H2O

Use the coefficient to convert mole of what you have to mols of what you want, then g = mols x molar mass

To calculate the grams of the product produced by the complete reaction, you need to use the molar mass of each reactant and product involved in the reaction.

First, let's determine the molar mass of CaCO3 (calcium carbonate), HCl (hydrochloric acid), CaCl2 (calcium chloride), CO2 (carbon dioxide), and H2O (water):

- Molar mass of CaCO3: 40.08 g/mol (Ca: 40.08 g/mol, C: 12.01 g/mol, O: 16.00 g/mol)
- Molar mass of HCl: 36.46 g/mol (H: 1.01 g/mol, Cl: 35.45 g/mol)
- Molar mass of CaCl2: 110.98 g/mol (Ca: 40.08 g/mol, Cl: 35.45 g/mol x 2)
- Molar mass of CO2: 44.01 g/mol (C: 12.01 g/mol, O: 16.00 g/mol x 2)
- Molar mass of H2O: 18.02 g/mol (H: 1.01 g/mol x 2, O: 16.00 g/mol)

Now that we know the molar masses, we can set up the stoichiometric ratio between the reactant (CaCO3) and the product (CO2) in order to find the grams of the product produced.

From the balanced equation, we see that 1 mole of CaCO3 reacts to produce 1 mole of CO2. Therefore, the stoichiometric ratio is 1:1.

Given that you have 0.125 moles of CaCO3, you can directly convert moles to grams using the molar mass:

Grams of CaCO3 = moles x molar mass
Grams of CaCO3 = 0.125 mol x 40.08 g/mol = 5.01 grams of CaCO3

Since the stoichiometric ratio between CaCO3 and CO2 is 1:1, the grams of CO2 produced would also be 5.01 grams.

Hence, by the complete reaction of 0.125 mole of CaCO3, 5.01 grams of CO2 would be produced.