Carbonates react with acids to form a salt, water and carbon dioxide gas. When 25.4 g of calcium carbonate are reacted with sufficient hydrochloric acid, how many grams of calcium chloride will be produced?
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
mols CaCO3 = grams/molar mass
Convert mols CaCO3 to mols CaCl2 using the coefficients in the balanced equation.
g CaCl2 = mols x molar mass.
55g
To find the grams of calcium chloride produced, we need to calculate the amount of calcium carbonate that reacts with hydrochloric acid and then determine the molar ratio between calcium carbonate and calcium chloride.
First, let's calculate the number of moles of calcium carbonate using its molar mass.
The molar mass of calcium carbonate (CaCO3) can be calculated by adding the atomic masses of its constituent elements:
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol (3 oxygen atoms in carbonate)
Molar mass of CaCO3 = (1 × 40.08 g/mol) + (1 × 12.01 g/mol) + (3 × 16.00 g/mol)
= 40.08 g/mol + 12.01 g/mol + 48.00 g/mol
= 100.09 g/mol
Next, we convert the mass of calcium carbonate given (25.4 g) to moles using its molar mass:
Moles of CaCO3 = (Mass of CaCO3) / (Molar mass of CaCO3)
= 25.4 g / 100.09 g/mol
= 0.2537 mol
According to the balanced equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl):
CaCO3 + 2HCl → CaCl2 + H2O + CO2
The molar ratio between CaCO3 and CaCl2 is 1:1. This means that for every 1 mol of calcium carbonate, we get 1 mol of calcium chloride.
Therefore, the number of moles of calcium chloride produced is also 0.2537 mol.
Finally, we can convert the moles of calcium chloride to grams using its molar mass.
The molar mass of calcium chloride (CaCl2) can be calculated by adding the atomic masses of its constituent elements:
- Calcium (Ca): 40.08 g/mol
- Chlorine (Cl): 35.45 g/mol (2 chlorine atoms in calcium chloride)
Molar mass of CaCl2 = (1 × 40.08 g/mol) + (2 × 35.45 g/mol)
= 40.08 g/mol + 70.90 g/mol
= 110.98 g/mol
Grams of CaCl2 = (Moles of CaCl2) × (Molar mass of CaCl2)
= 0.2537 mol × 110.98 g/mol
= 28.63 g
Therefore, when 25.4 g of calcium carbonate are reacted with sufficient hydrochloric acid, approximately 28.63 grams of calcium chloride will be produced.