Question

In the equation given below, 100 g of CaCO₃ and 73 g of HCl are used to prepare 18 g of H₂O. If 300 g of CaCO₃ and 146 g of HCl are used, then how many grams of H₂O is produced?
CaCO₃ + 2HCl → CaCl₂ + H₂O+CO₂

• A. 54
• B. 36
• C. 300
• D. 146

Answer 1

The Correct Option is B

The balanced chemical equation is: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

First, let's find the molar masses of the relevant substances:

• Molar mass of CaCO₃ = 40.08 (Ca) + 12.01 (C) + 3 x 16.00 (O) ≈ 100 g/mol
• Molar mass of HCl = 1.01 (H) + 35.45 (Cl) ≈ 36.5 g/mol
• Molar mass of H₂O = 2 x 1.01 (H) + 16.00 (O) ≈ 18 g/mol

From the balanced equation, the stoichiometric ratio is: 1 mole of CaCO₃ reacts with 2 moles of HCl to produce 1 mole of H₂O. In terms of mass: 100 g of CaCO₃ reacts with 2 x 36.5 g = 73 g of HCl to produce 1 x 18 g = 18 g of H₂O. This confirms the information given in the first part of the problem.

Now, consider the second case where 300 g of CaCO₃ and 146 g of HCl are used. We need to find the limiting reactant.

Convert the given masses to moles:

• Moles of CaCO₃ = 300 g / 100 g/mol = 3.0 mol
• Moles of HCl = 146 g / 36.5 g/mol = 4.0 mol

Determine the limiting reactant based on the stoichiometric ratio (1 mole CaCO₃ : 2 moles HCl):

• To react completely with 3.0 mol of CaCO₃, we need 3.0 x 2 = 6.0 mol of HCl. We only have 4.0 mol of HCl.
• To react completely with 4.0 mol of HCl, we need 4.0 / 2 = 2.0 mol of CaCO₃. We have 3.0 mol of CaCO₃, which is more than enough.

Therefore, HCl is the limiting reactant.

The amount of product (H₂O) formed is determined by the amount of the limiting reactant (HCl). From the balanced equation, 2 moles of HCl produce 1 mole of H₂O.

Calculate the moles of H₂O produced using the moles of the limiting reactant (HCl): Moles of H₂O = (4.0 mol HCl) x (1 mol H₂O / 2 mol HCl) = 2.0 mol H₂O

Convert the moles of H₂O to grams: Mass of H₂O = Moles of H₂O x Molar mass of H₂O Mass of H₂O = 2.0 mol x 18 g/mol = 36 g

The correct option is (B): 36