When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.

CaCO3(s)+2HCl(aq)=CaCl2(aq)+H2O(l)+CO2(g)
How many grams of calcium chloride will be produced when 27.0 g of calcium carbonate are combined with 11.0 g of hydrochloric acid?
Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?

mols CaCO3 = grams/molar mass = ?

mols HCl = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols CaCO3 to mols CaCl2.
Do the same to convert mols HCl to mols CaCl2.
It is likely that the two values will not agree which means one of them is not right. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.

Now convert the smaller value of mols CaCl2 to grams. grams CaCl2 = mols CaCl2 x molar mass CaCl2.

For the excess reagent.
LR = limiting reagent
OR = other reagent; i.e., the non-limiting reagent.

In a process similar to converting mols HCl and CaCO3 to mols CaCl2, convert mols LR to mols of the OR. That will give you mols the OR used by the LR. Then mols OR initially - mols OR used = mols OR left after the reaction is complete.

To determine the grams of calcium chloride produced and identify the excess reactant, we need to follow these steps:

Step 1: Calculate the moles of each reactant:
- Convert the mass of calcium carbonate to moles by dividing by its molar mass (40.08 g/mol for CaCO3):
Moles of CaCO3 = 27.0 g / 40.08 g/mol = 0.674 mol

- Convert the mass of hydrochloric acid to moles by dividing by its molar mass (36.46 g/mol for HCl):
Moles of HCl = 11.0 g / 36.46 g/mol = 0.302 mol

Step 2: Identify the limiting reactant:
- The limiting reactant is the one that is completely consumed in the reaction, thereby limiting the amount of product formed.
- To find the limiting reactant, we need to compare the stoichiometric ratio between the reactants and the desired product.
- The balanced equation shows a 1:2 ratio between CaCO3 and CaCl2. Therefore, if we have less than twice the moles of CaCO3 than the moles of HCl, CaCO3 is the limiting reactant.

Since we have 0.674 mol of CaCO3 and 0.302 mol of HCl, the HCl is in excess.

Step 3: Calculate the grams of calcium chloride produced:
- The balanced equation tells us that for every 1 mole of CaCO3, we get 1 mole of CaCl2.
- Calculate the moles of CaCl2 produced using the moles of CaCO3 (which is equal to CaCl2):
Moles of CaCl2 produced = 0.674 mol

- Convert the moles of CaCl2 to grams by multiplying by its molar mass (110.98 g/mol for CaCl2):
Grams of CaCl2 produced = 0.674 mol * 110.98 g/mol = 74.89 g

Therefore, 74.89 grams of calcium chloride will be produced when 27.0 grams of calcium carbonate react with 11.0 grams of hydrochloric acid.

Step 4: Calculate the grams of excess reactant remaining:
- To find the grams of excess reactant remaining, we need to determine how much of it was not used up.
- Start with the moles of the excess reactant (HCl):
Moles of HCl remaining = Moles of HCl initially - Moles of HCl consumed
= 0.302 mol - 0.674 mol (from the stoichiometry)

- Convert the moles of HCl remaining to grams by multiplying by its molar mass (36.46 g/mol for HCl):
Grams of HCl remaining = 0.302 mol * 36.46 g/mol = 11.00 g

So, 11.00 grams of hydrochloric acid will remain after the reaction is complete.