Problem 138 - General Chemistry

Question

Barium carbonate is the source of barium compounds. It is produced in an aqueous precipitation reaction from barium sulfide and sodium carbonate. (Barium sulfide is a soluble compound obtained by heating the mineral barite, which is barium sulfate, with carbon.) What are the molecular equation and net ionic equation for the precipitation reaction? A solution containing \(33.9 \mathrm{~g}\) of barium sulfide requires \(21.2 \mathrm{~g}\) of sodium carbonate to react completely with it, and \(15.6 \mathrm{~g}\) of sodium sulfide is produced in addition to whatever barium carbonate is obtained. How many grams of barium sulfide are required to produce \(5.00\) tons of barium carbonate? (One ton equals 2000 pounds.)

Step-by-Step Solution

Answer

8.59 tons of barium sulfide are needed to produce 5 tons of barium carbonate.

Step 1: Write the molecular equation

The reaction between barium sulfide (\(\text{BaS}\)) and sodium carbonate (\(\text{Na}_2\text{CO}_3\)) forms barium carbonate (\(\text{BaCO}_3\)) and sodium sulfide (\(\text{Na}_2\text{S}\)). The molecular equation is:\[\text{BaS (aq) + Na}_2\text{CO}_3\text{ (aq) } \rightarrow \text{BaCO}_3\text{ (s) + Na}_2\text{S (aq)}\]

Step 2: Write the net ionic equation

The net ionic equation focuses only on the ions that form the precipitate. Barium ions (\(\text{Ba}^{2+}\)) and carbonate ions (\(\text{CO}_3^{2-}\)) form barium carbonate:\[\text{Ba}^{2+}\text{ (aq) } + \text{CO}_3^{2-}\text{ (aq) } \rightarrow \text{BaCO}_3\text{ (s)}\]

Step 3: Convert tons of product to grams

5 tons of barium carbonate is equal to \(5 \times 2000 \text{ pounds }\times 453.6 \text{ grams per pound} = 4,536,000 \text{ grams of }\text{BaCO}_3\).

Step 4: Calculate moles of barium carbonate

The molar mass of barium carbonate is calculated as follows:\[\text{BaCO}_3: \text{Ba} = 137.33, \text{C} = 12.01, \text{O} = 16.00 \times 3\] Thus, \[\text{Molar mass of BaCO}_3 = 137.33 + 12.01 + 48.00 = 197.34 \text{ g/mol}\]Convert grams to moles:\[\frac{4,536,000 \text{ g}}{197.34 \text{ g/mol}} = 22,986.25 \text{ mol}\]

Step 5: Find moles of barium sulfide

Assuming a 1:1 molar ratio from the chemical equation, the moles of barium sulfide (\(\text{BaS}\)) needed is the same as the moles of barium carbonate, 22,986.25 mol.

Step 6: Convert moles of barium sulfide to grams

Calculate the molar mass of barium sulfide:\[\text{BaS:}\, \text{Ba} = 137.33, \text{S} = 32.06\]Thus,\[\text{Molar mass of BaS} = 137.33 + 32.06 = 169.39 \text{ g/mol}\]Convert moles to grams:\[22,986.25 \text{ mol} \times 169.39 \text{ g/mol} = 3,895,762.39 \text{ grams}\]

Step 7: Convert grams to tons

Convert the grams of barium sulfide to tons:\[\frac{3,895,762.39 \text{ grams}}{453,600 \text{ grams/ton}} \approx 8.59 \text{ tons}\]

Key Concepts

Molecular Equation

In a precipitation reaction, the molecular equation is a crucial initial step. It shows all reactants and products in their chemical forms. This includes compounds in aqueous solutions and the solid precipitate formed. For the reaction between barium sulfide (\(\text{BaS}\)) and sodium carbonate (\(\text{Na}_2\text{CO}_3\)), the molecular equation would be:
\[ \text{BaS} (aq) + \text{Na}_2\text{CO}_3 (aq) \rightarrow \text{BaCO}_3 (s) + \text{Na}_2\text{S} (aq) \]This equation indicates that when barium sulfide and sodium carbonate react in an aqueous medium, barium carbonate precipitate and sodium sulfide solution are produced.
Molecular equations are useful because they show the complete chemical formulas of the reactants and products, hence providing an overall picture of the reaction.

Net Ionic Equation

The net ionic equation simplifies the chemical equation by focusing solely on the species that participate in the formation of a precipitate. This enhances our understanding of the actual chemical change occurring in the reaction.
In the given problem, the ions responsible for the formation of barium carbonate (\(\text{BaCO}_3\)) precipitate are the barium ions (\(\text{Ba}^{2+}\)) and carbonate ions (\(\text{CO}_3^{2-}\)). The net ionic equation becomes:
\[ \text{Ba}^{2+}(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{BaCO}_3(s) \]
This equation omits the spectator ions, which in this case are the sodium ions (\(\text{Na}^+\)) and sulfide ions (\(\text{S}^{2-}\)), as they remain unchanged in the reaction.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It allows us to predict the amounts of substances consumed and produced in the reaction, based on balanced chemical equations.
For this reaction, the stoichiometric idea assumes a 1:1 molar relationship between barium sulfide and barium carbonate. Given that every molecule of \(\text{BaS}\) reacts with one unit of \(\text{Na}_2\text{CO}_3\) to produce one unit of \(\text{BaCO}_3\), we can deduce that the moles of \(\text{BaS}\) used are equal to the moles of \(\text{BaCO}_3\) produced.
This direct stoichiometric relationship is valuable for calculating quantities, and it’s precisely how we can determine how much barium sulfide is required to produce a given amount of barium carbonate.

Molar Mass Calculation

The calculation of molar mass is an important step in solving many chemical problems, including precipitation reactions. It involves summing up the atomic masses of all elements in a compound’s formula.
For barium carbonate (\(\text{BaCO}_3\)), we calculate the molar mass by adding the atomic weights of barium (\text{Ba}\, = 137.33), carbon (\text{C}\, = 12.01), and three oxygen (\text{O}\, = 16.00 each) atoms:
\[ \text{Molar mass of } \text{BaCO}_3 = 137.33 + 12.01 + 3 \times 16.00 = 197.34 \text{ g/mol} \]
Understanding molar mass allows us to convert between the mass of a substance and the number of moles. This conversion is necessary for further stoichiometric calculations, such as determining how much reactant is needed or how much product can be formed.