When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
CaCO3 +2HCl yields CaCl2 +H2O +CO2
How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate are combined with 13.0 g of hydrochloric acid?
Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
Convert 29.0g CaCO3 to mols. mol = grams/molar mass.
Convert 13.0g HCl to mols.
Using the coefficients in the balanced equation, convert mols CaCO3 to mols CaCl2.
Do the same for mols HCl to mols CaCl2.
It is likely that the two values will not agree which means one of them is wrong; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. The other reagent is the one in excess.
g CaCl2 produced is the smaller value for mols x molar mass CaCl2 = ?
Now that you know the excess and limiting reagents, use the coefficients to convert mols of the limiting reagent used to mols excess reagent used, subtract mols excess initially - mols excess reagent used = mols excess reagent remaining. The convert that to grams by g = mols x molar mass.
To determine the grams of calcium chloride produced and the excess reactant, we need to follow these steps:
Step 1: Convert the given masses of calcium carbonate and hydrochloric acid to moles. To do this, we will use the molar mass of each compound.
The molar mass of CaCO3 (calcium carbonate) is:
40.08 g/mol (atomic mass of Ca) + 12.01 g/mol (atomic mass of C) + 3(16.00 g/mol) (atomic mass of O) = 100.09 g/mol
Moles of calcium carbonate = mass / molar mass = 29.0 g / 100.09 g/mol = 0.289 mol
The molar mass of HCl (hydrochloric acid) is:
1.01 g/mol (atomic mass of H) + 35.45 g/mol (atomic mass of Cl) = 36.46 g/mol
Moles of hydrochloric acid = mass / molar mass = 13.0 g / 36.46 g/mol = 0.357 mol
Step 2: Determine the mole ratio between calcium carbonate and calcium chloride from the balanced chemical equation. According to the equation, 1 mol of CaCO3 reacts with 1 mol of CaCl2.
Step 3: Using the mole ratio, calculate the moles of CaCl2 produced. Since the mole ratio is 1:1, the moles of CaCl2 produced will be the same as the moles of calcium carbonate.
Moles of calcium chloride = 0.289 mol
Step 4: Convert the moles of calcium chloride to grams. We will use the molar mass of CaCl2, which is:
40.08 g/mol (atomic mass of Ca) + 2(35.45 g/mol) (atomic mass of Cl) = 110.98 g/mol
Grams of calcium chloride = moles × molar mass = 0.289 mol × 110.98 g/mol = 31.58 g
Therefore, 31.58 grams of calcium chloride will be produced.
Step 5: Determine the excess reactant by comparing the amounts of moles used in the reaction. The reactant with the higher mole amount is the excess reactant.
Since the mole ratio between calcium carbonate (CaCO3) and hydrochloric acid (HCl) is 1:2 (according to the balanced equation), for every 1 mole of CaCO3, we need 2 moles of HCl.
Moles of hydrochloric acid required (based on the balanced equation): 2 × moles of calcium carbonate = 2 × 0.289 mol = 0.578 mol
The initial moles of hydrochloric acid given were 0.357 mol. Since this value is lower than the required 0.578 mol, hydrochloric acid is the limiting reactant. Calcium carbonate is in excess.
Step 6: Calculate the remaining grams of the excess reactant. We can find this by subtracting the moles used in the reaction from the initial moles given, and then converting it to grams.
Moles of hydrochloric acid remaining = moles of hydrochloric acid initially - moles of hydrochloric acid used in the reaction
= 0.357 mol - (0.289 mol × 2) = 0.357 mol - 0.578 mol = -0.221 mol (negative value indicates it is completely used up)
Grams of hydrochloric acid remaining = moles × molar mass = -0.221 mol × 36.46 g/mol ≈ -8.06 g (negative value indicates it is completely used up)
Therefore, there will be no grams of hydrochloric acid remaining after the reaction is complete.