Is the mass of calcium chloride the same as the mass of anhydrous calcium chloride? I'm confused.

Question

Answer the following questions based a reaction between 80.0 g of sodium hydroxide and 90.0 g of anhydrous calcium chloride.

Write a balanced chemical equation representing this reaction.

How many moles of the product compound containing calcium will be made from the given mass of anhydrous calcium chloride?

Answer 1

Yes, anhydrous means without any water attached, so the molar mass of CaCl2 is meant at the anhydrous salt, and is 111 g / mol.

2NaOH(aq) + CaCl₂(aq) ==> 2NaCl(aq) + Ca(OH)₂(s) ... balanced equation

Finding the limiting reactant. One easy way is to divide the moles of each reactant by the corresponding coefficient in the balanced equation.

NaOH: 80.0 g x 1 mol / 40 g = 2 mols (÷2 -> 1)

CaCl₂: 90.0 g x 1 mol / 111 g = 0.81 mols (÷1 -> 0.81)

Limiting reactant is CaCl₂ b/c 0.81 is less than 1.

The product containing Ca is the Ca(OH)₂, with a molar mass of 74.1 g/mol

Finding the amount of Ca(OH)₂ formed:

0.81 mols Ca(OH)₂ x 1 mol Ca(OH)₂ / mol CaCl₂ x 74.1 g / mol = 60.1 g Ca(OH)₂