CaCO3(s)+2HCl(aq)>CaCl2(aq)+H2O(l)+CO2(g)
how many grams of calcium chloride will be produced when 30 g of calcium carbonate are combined with 13 g of hydrochloric acid?
Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?
CaCO3(s)+2HCl(aq)>CaCl2(aq)+H2O(l)+CO2(g)
I will estimate all of these numbers; you need to go through and recalculate each.
mols CaCO3 = grams/molar mass = approx 0.30
mols HCl = grams/molar mass = approx 0.36
Convert mols CaCO3 to mols CaCl2 using the coefficients in the balanced equation.
CaCO3 to mols CaCl2 = approx 0.30
HCl to mols CaCl2 = approx 0.18
In limiting reagent (LR) problems the correct value is ALWAYS the smaller value and the reagent producing that value is the LR. So HCl is the LR and CaCO3 is in excess.
mols CaCl2 = 0.18 x molar mass CaCl2 = ?
How much of the CaCO3 is left unreacted? Using the coeffients again, convert mols HCl used to mols CaCO3. That's 0.36 mols HCl x (1 mol CaCO3/2 mols HCl) = 0.18 mols CaCO3. You started with approx 0.30 and used approx 0.18 so you have approx 0.12 mols CaCO3 remaining unreacted. Converted to grams that is g = mols x molar mass = ?
To determine the grams of calcium chloride produced, we need to calculate the moles of calcium carbonate first.
1. Calculate the moles of calcium carbonate (CaCO3):
Moles = Mass / Molar mass
Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 3(16.00 g/mol) (O) = 100.09 g/mol
Moles of CaCO3 = 30 g / 100.09 g/mol ≈ 0.2999 mol (approximately 0.30 mol)
From the balanced equation, we can see that the stoichiometric ratio between CaCO3 and CaCl2 is 1:1. Therefore, the moles of calcium chloride produced will also be 0.30 mol.
2. Convert the moles of calcium chloride to grams:
Mass = Moles × Molar mass
Molar mass of CaCl2 = 40.08 g/mol (Ca) + 2(35.45 g/mol) (Cl) = 110.98 g/mol
Mass of CaCl2 = 0.30 mol × 110.98 g/mol ≈ 33.30 g (approximately 33.3 g)
Therefore, approximately 33.3 grams of calcium chloride will be produced.
Now, let's determine the excess reactant:
3. Calculate the moles of hydrochloric acid (HCl):
Moles = Mass / Molar mass
Molar mass of HCl = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46 g/mol
Moles of HCl = 13 g / 36.46 g/mol ≈ 0.3564 mol (approximately 0.356 mol)
From the balanced equation, we can see the stoichiometric ratio between CaCO3 and HCl is 1:2. Therefore, we need twice as many moles of HCl as CaCO3. Since 0.30 mol of CaCO3 is used, we need 2 × 0.30 mol = 0.60 mol of HCl.
Since we only have 0.356 mol of HCl, the HCl is in excess.
4. Calculate the remaining mass of HCl:
Moles of excess HCl = Moles of HCl initially - Moles of HCl used
Moles of excess HCl = 0.356 mol - 0.60 mol = -0.244 mol
Since we have a negative value, it means there will be no HCl remaining after the reaction is complete.
Therefore, the excess reactant is hydrochloric acid (HCl), and there will be no grams of HCl remaining after the reaction is complete.
To determine the grams of calcium chloride produced and the excess reactant, you need to follow these steps:
1. Write and balance the chemical equation:
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
2. Calculate the number of moles of each reactant:
The molar mass of CaCO3 is: 40.08 g/mol (calcium) + 12.01 g/mol (carbon) + 3(16.00 g/mol) (oxygen) = 100.09 g/mol
Moles of CaCO3 = 30 g / 100.09 g/mol = 0.2998 mol
The molar mass of HCl is: 1.01 g/mol (hydrogen) + 35.45 g/mol (chlorine) = 36.46 g/mol
Moles of HCl = 13 g / 36.46 g/mol = 0.3562 mol
3. Determine the limiting reactant:
The reaction equation shows a 1:2 ratio between CaCO3 and HCl. In this case, the reaction requires twice as many moles of HCl as CaCO3.
However, we have 0.2998 mol of CaCO3 and 0.3562 mol of HCl. Therefore, CaCO3 is limiting, and HCl is in excess.
4. Calculate the moles of products formed:
From the equation, we see that 1 mole of CaCO3 produces 1 mole of CaCl2.
Moles of CaCl2 produced = 0.2998 mol of CaCO3
5. Convert the moles of CaCl2 to grams:
The molar mass of CaCl2 is: 40.08 g/mol (calcium) + 2(35.45 g/mol) (chlorine) = 110.98 g/mol
Grams of CaCl2 = Moles of CaCl2 × Molar mass of CaCl2
= 0.2998 mol × 110.98 g/mol ≈ 33.25 g
So, approximately 33.25 grams of calcium chloride will be produced when 30 grams of calcium carbonate and 13 grams of hydrochloric acid react.
To determine the grams of excess reactant remaining, we need to calculate the moles of the excess reactant (HCl) and convert it to grams:
Moles of HCl remaining = Moles of HCl initially - Moles of HCl used in the reaction
= (0.3562 mol HCl initially) - (0.2998 mol HCl used)
= 0.0564 mol HCl remaining
Grams of HCl remaining = Moles of HCl remaining × Molar mass of HCl
= 0.0564 mol × 36.46 g/mol ≈ 2.05 g
Therefore, after the reaction is complete, approximately 2.05 grams of hydrochloric acid will remain as the excess reactant.