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

CaCO3 + 2HCl -> CaCl2 + H2O + CO2
How many grams of calcium chloride will be produced when 27.0 g of calcium carbonate are combined with 14.0 g of hydrochloric acid?

Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?

?g of CaCO3 or HCl

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

Convert 27.0 g CaCO3 to mols.
Convert 14.0 g HCl to mols.

Use the coefficients to convert mols CaCO3 to mols CaCl2.
Do the same for HCl.
It is likely that the two values will not be the same (mols CaCl2 that is) and one of them must be wrong. The correct answer in limiting reagent is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Using the smaller value convert mols CaCl2 to grams. g = mols x molar mass.

To find the amount of excess reagent unreacted do this.
Using the coefficients convert mols of the limiting reagent to mols of the "other" reagent. That gives the amount of the non-limiting reagent USED. Convert to grams and subtract from the initial amount. The difference is the amount of excess reagent unreacted.
Post your work if you get stuck.

I got 0.270 mol CaCO3, 0.384 mol HCl, and 0.270 mol CaCl2. The LR is HCl. I don't know what to do next.

You're ok for part of it. 0.270 mol CaCO3 is right. 0.384 mol HCl is right. But you will not form 0.170 mol CaCl2. You have LR of HCl but you must have arrived at that answer incorrectly.

Convert mols CaCO3 to mols CaCl2 by
0.270 x (1 mol CaCl2/1 mol CaCO3) = 0.270.
Convert mols HCl to mols CaCl2 by
0.384 mols HCl x (1 mol CaCl2/2 mol HCl) = 0.192 mols CaCl2. Thus mols CaCl2 by HCl is the smaller value of 0.279 and 0.192 and 0.192 wins as the LR.
So you will form 0.385 x 1/2 = 0.192 mols CaCl2. Convert that to grams by g = mols x molar mass CaCl2.

To find how much CaCO3 is left (the excess reagent), convert mols HCl to mols CaCO3. That's 0.384 x (1 mol CaCO3/2 mol HCl) = 0.384 x 1/2 = 0.192 mols CaCO3 used. Grams CaCO3 used = 0.192 x molar mass CaCO3 = approximately 19 g.
You started with 27.0, you've used about 19 so you must have 27-19 = about 8 g CaCO3 left over.

To determine the number of grams of calcium chloride produced and which reactant is in excess, we need to follow these steps:

Step 1: Calculate the number of moles of each reactant using their respective molar masses.

The molar mass of calcium carbonate (CaCO3) is:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (3 oxygen atoms in CaCO3)

So, the total molar mass of CaCO3 is: 40.08 + 12.01 + (16.00 x 3) = 100.09 g/mol

To calculate the number of moles of CaCO3, divide the mass (27.0 g) by its molar mass:
Number of moles of CaCO3 = 27.0 g / 100.09 g/mol = 0.269 mol

The molar mass of hydrochloric acid (HCl) is:
H: 1.01 g/mol
Cl: 35.45 g/mol

So, the total molar mass of HCl is: 1.01 + 35.45 = 36.46 g/mol

To calculate the number of moles of HCl, divide the mass (14.0 g) by its molar mass:
Number of moles of HCl = 14.0 g / 36.46 g/mol = 0.384 mol

Step 2: Use the balanced equation to determine the stoichiometric ratio between moles of CaCO3 and CaCl2.

From the balanced equation: CaCO3 + 2HCl -> CaCl2 + H2O + CO2
We can observe that 1 mole of CaCO3 reacts with 1 mole of CaCl2.

Step 3: Determine the limiting reactant.

The limiting reactant is the one that is completely consumed and determines the amount of product formed. It can be determined by comparing the moles of each reactant.

In this case, 1 mole of CaCO3 reacts with 1 mole of CaCl2. Therefore, the limiting reactant is the one with fewer moles.

Since we have 0.269 moles of CaCO3 and 0.384 moles of HCl, the limiting reactant is CaCO3.

Step 4: Calculate the grams of calcium chloride produced.

Since we know that the stoichiometric ratio is 1:1 between CaCO3 and CaCl2, the number of moles of CaCl2 produced will be the same as the number of moles of CaCO3.

Therefore, the number of moles of CaCl2 produced is 0.269 mol.

To calculate the grams of calcium chloride produced, multiply the number of moles by its molar mass (Ca: 40.08 g/mol, Cl: 35.45 g/mol):
Mass of CaCl2 produced = 0.269 mol x (40.08 g/mol + 2 * 35.45 g/mol) = 27.907 g

So, 27.907 grams of calcium chloride will be produced.

Step 5: Determine the excess reactant and calculate the remaining grams.

The excess reactant is the one that is not completely consumed and will be present in excess after the reaction is complete.

Since CaCO3 is the limiting reactant, HCl is in excess.

To calculate the remaining grams of HCl, subtract the grams used in the reaction from the initial amount:

Remaining grams of HCl = Initial grams of HCl - grams used in reaction
Remaining grams of HCl = 14.0 g - (0.269 mol x 36.46 g/mol) = 3.089 g

Therefore, approximately 3.089 grams of hydrochloric acid will remain after the reaction is complete.