Question

Calculate the mass of the precipitate formed when \(2.27 \mathrm{~L}\) of \(0.0820 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) is mixed with \(3.06 \mathrm{~L}\) of \(0.0664 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)

Short answer

The mass of the precipitate is 43.44 g.

Step-by-step solution

Identify the Reaction

When barium hydroxide, \(\text{Ba(OH)}_2\), is mixed with sodium sulfate, \(\text{Na}_2\text{SO}_4\), a precipitation reaction occurs, forming barium sulfate, \(\text{BaSO}_4\), as a solid precipitate, according to the reaction: \(\text{Ba(OH)}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2\text{NaOH}\).

Calculate Moles of Reactants

First, calculate the number of moles of \(\text{Ba(OH)}_2\) by multiplying its concentration by its volume: \(\text{Moles of } \text{Ba(OH)}_2 = 2.27 \text{ L} \times 0.0820 \text{ M} = 0.18614 \text{ moles}\). Similarly, calculate the moles of \(\text{Na}_2\text{SO}_4\) by multiplying its concentration by its volume: \(\text{Moles of } \text{Na}_2\text{SO}_4 = 3.06 \text{ L} \times 0.0664 \text{ M} = 0.203184 \text{ moles}\).

Determine Limiting Reactant

In the reaction, \(1\) mole of \(\text{Ba(OH)}_2\) reacts with \(1\) mole of \(\text{Na}_2\text{SO}_4\) to form \(1\) mole of \(\text{BaSO}_4\). Therefore, the reactant with fewer moles is the limiting reactant. Since \(0.18614 \text{ moles of Ba(OH)}_2\) is less than \(0.203184 \text{ moles of Na}_2\text{SO}_4\), \(\text{Ba(OH)}_2\) is the limiting reactant.

Calculate Moles of Precipitate

Because \(\text{Ba(OH)}_2\) is the limiting reactant, the moles of \(\text{BaSO}_4\) formed will be equal to the moles of \(\text{Ba(OH)}_2\), which is \(0.18614 \text{ moles}\).

Convert Moles to Mass

To find the mass of \(\text{BaSO}_4\), multiply its moles by its molar mass \(\text{(233.39 g/mol)}\): \(\text{Mass} = 0.18614 \text{ moles} \times 233.39 \text{ g/mol} = 43.44 \text{ g}\).

Key concepts

Barium Hydroxide

Barium hydroxide, represented as \( ext{Ba(OH)}_2\), is an inorganic compound known for its role in chemistry as a strong base. It dissociates in water to produce barium ions \((\text{Ba}^{2+})\) and hydroxide ions \((\text{OH}^- )\).When barium hydroxide is involved in a chemical reaction, it often participates in precipitation reactions. A precipitation reaction occurs when two aqueous solutions combine to form an insoluble solid, known as a precipitate.In the case of the exercise, mixing barium hydroxide with sodium sulfate \((\text{Na}_2 \text{SO}_4)\) leads to the formation of a solid precipitate, barium sulfate \((\text{BaSO}_4 \downarrow)\). This is illustrated by the reaction equation: \[\text{Ba(OH)}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2\text{NaOH}\]By carrying out this reaction, we can identify essential topics such as the identification of reactants and the formation of products. Here, barium hydroxide plays a crucial role in determining the product that precipitates from the solution.

Limiting Reactant

The concept of a limiting reactant is fundamental when dealing with reactions in stoichiometry.A limiting reactant is the substance that is entirely consumed first during a chemical reaction, determining the amount of product formed.To identify the limiting reactant, compare the mole ratio of the reactants to the ratio required by the balanced chemical equation.In our specific exercise, the balanced equation shows \(1\) mole of \(\text{Ba(OH)}_2\) reacts with \(1\) mole of \(\text{Na}_2\text{SO}_4\) to yield \(1\) mole of \(\text{BaSO}_4\).In the given scenario:

• We have \(0.18614\) moles of \(\text{Ba(OH)}_2\).
• We have \(0.203184\) moles of \(\text{Na}_2\text{SO}_4\).

Since \(\text{Ba(OH)}_2\) has fewer moles available than \(\text{Na}_2\text{SO}_4\), it is consumed first, making \(\text{Ba(OH)}_2\) the limiting reactant.Identifying the limiting reactant allows us to predict the exact amount of the precipitate \(\text{BaSO}_4\) formed.

Molar Mass

Molar mass is another critical concept in these types of calculations. It refers to the mass of one mole of a given substance, typically expressed in grams per mole (g/mol). For precise calculations, it is crucial to determine the molar mass of the products and reactants involved in a chemical reaction.Using the periodic table, you can find the atomic masses and derive the molar mass of compounds. For example, the molar mass of barium sulfate \((\text{BaSO}_4)\), as given in the step-by-step solution, is \(233.39 \text{ g/mol}\).In the exercise, the importance of molar mass arises when converting the moles of \(\text{BaSO}_4\) into grams to determine the precipitate's mass. To find the mass, multiply the moles by the molar mass:\[\text{Mass} = \text{Moles} \times \text{Molar Mass}\]For the precipitate \(\text{BaSO}_4\), this calculation gives:\[0.18614 \text{ moles} \times 233.39 \text{ g/mol} = 43.44 \text{ g}\]Understanding molar mass ensures accurate predictions and measurements in chemical reactions, making it an indispensable tool for any chemist.