Double Displacement Reactions with Hydrates

Double Displacement Reactions with Hydrates.

Hydrates are substances that include water in their formulas. The water is not actually part of the chemical substance and this is reflected in the way the formula is written.

For example:copper(II) sulfate pentahydrate

CuSO4⋅5H2O

This formula means that for every CuSO4formula unit in the piece of this substance you have, there are also five water molecules. The substance is not wet, it appears and feels dry. There are some hydrates that have a wet appearance, but most appear perfectly dry to the eye and to the touch.

The dot IS NOT a multiplication sign. It is just a way of showing that for one formula unit of copper(II) sulfate there are five molecules of water included in its crystalline structure.

Please understand, not all samples of CuSO4 are hydrates. CuSO4 without attached water molecules is referred to as anhydrous copper(II) sulfate (or, simply, copper(II) sulfate). A bottle of copper(II) sulfate crystals will indicate if it is anhydrous or a hydrate.

The molar mass of hydrated compounds must include the mass of the water.

Ex: CuSO4⋅5H2O

Naming hydrates.

Use the prefixes for numbers 1 through 10 (mono-, di-, tri-, tetra-, penta-, hexa-, hepta-, octa-, nona-, deca-) to indicate the number of water molecules present in the compound formula.

Ex: CuSO4⋅5H2O is named: copper(II) sulfate pentahydrate

Ex: CaCl2⋅2H2O is named: calcium chloride dihydrate

Name these hydrates:

MgSO4 ⋅ 7H2O______

Ca(C2H3O2)2⋅H2O______

Ba(OH)2⋅8H2O______

CoCl2⋅6H2O______

ZnSO4⋅7H2O______

Stoichiometry with Hydrates.

To balance an equation with hydrate compounds, simply ignore the presence of the water molecules, determine the products, and balance the reaction. Then include all the hydrate water molecules as a separate product. If water is also produced in the reaction, simply add up all the water molecules as a single separate product.

Finish the precipitation reactions below, balance them, and determine the products:

Ba(OH)2⋅8H2O + HNO3 

CoCl2⋅6H2O + Na2S 

Lab: Can You Make 2.00 Grams of a Compound?

Prelab: Prepare 2.00 grams of precipitate by reacting barium chloride dihydrate with silver nitrate.

NOTE: You are not using a limiting reactant in this lab. You are computing the precise amounts of two reactants needed to make 2.00 grams of a product. Both reactants, ideally, will be completely consumed in the reaction. Remember, use grams  moles  moles  grams to compute masses of each reactant.

1.Write the balanced equation and identify the product.

2.Calculate molar masses for the reactants and the product.

3.How much barium chloride dihydrate is required to make 2.00 grams of the product?

4.How much silver nitrate is required to make 2.00 grams of the product?