# A mixture of 0.437 g of BaCl2 cdot 2H2O and 0.284 g Na2SO4 is added to water. What is the yield...

## Question:

A mixture of 0.437 g of {eq}BaCl_2 \cdot 2H_2O {/eq} and 0.284 g {eq}Na_2SO_4 {/eq} is added to water. What is the yield of {eq}BaSO_4 {/eq} precipitate (in grams)?

## Limiting Reactant:

The theoretical yield of a reaction is based on the mass available of the limiting reactant. This reactant is the species that is consumed first based on the mass of the reactants, the molar mass of each, and their molar ratios.

## Answer and Explanation: 1

The reaction of these two salts follows the double-displacement reaction of:

{eq}\rm BaCl_2 \cdot 2H_2O + Na_2SO_4 \to BaSO_4 + 2NaCl + 2H_2O {/eq}

This means that barium chloride hydrate and sodium sulfate react on a 1 : 1 molar basis to produce 1 mole of barium sulfate. The mass of each species can be converted to and from a quantity of moles using the molar mass of each species. The molar mass of {eq}BaCl_2 \cdot 2H_2O {/eq} is 244.26 g/mol. The molar mass of {eq}Na_2SO_4 {/eq} is 142.04 g/mol. The molar mass of {eq}BaSO_4 {/eq} is 233.39 g/mol.

To find the limiting reactant, calculate the mass of the second reactant required to react with the mass of the first reactant:

{eq}0.437\ g\ \mathrm{BaCl_2}\cdot 2\mathrm{H_2O} \left(\dfrac{1\ mol\ \mathrm{BaCl_2}\cdot 2\mathrm{H_2O}}{244.26\ g\ \mathrm{BaCl_2}\cdot 2\mathrm{H_2O}}\right) \left(\dfrac{1\ mol\ \mathrm{Na_2SO_4}}{1\ mol\ \mathrm{BaCl_2}\cdot 2\mathrm{H_2O}}\right) \left(\dfrac{142.04\ g\ \mathrm{Na_2SO_4}}{1\ mol\ \mathrm{Na_2SO_4}}\right) = 0.254\ g\ \mathrm{Na_2SO_4} {/eq}

Since only 0.254 g of sodium sulfate is necessary to react with this mass of the hydrate but 0.284 g were available, this makes barium chloride dihydrate the limiting reactant.

The theoretical yield of barium sulfate is based on the mass of the limiting reactant:

{eq}0.437\ g\ \mathrm{BaCl_2}\cdot 2\mathrm{H_2O} \left(\dfrac{1\ mol\ \mathrm{BaCl_2}\cdot\mathrm{H_2O}}{244.26\ g\ \mathrm{BaCl_2}\cdot 2\mathrm{H_2O}}\right) \left(\dfrac{1\ mol\ \mathrm{BaSO_4}}{1\ mol\ \mathrm{BaCl_2}\cdot 2\mathrm{H_2O}}\right) \left(\dfrac{233.39\ g\ \mathrm{BaSO_4}}{1\ mol\ \mathrm{BaSO_4}}\right) = \boxed{\: 0.418\ g\ \mathrm{BaSO_4} \:} {/eq}

#### Learn more about this topic:

from

Chapter 9 / Lesson 7Learn the definition of a limiting reactant, the formula, and how to determine a limiting reactant. See examples of limiting reactants and their problems.