Calculating Product Mass In Silver Nitrate And Barium Sulfate Reaction

July 13, 2025 by Scholario Team 71 views

**Understanding the Reaction of Silver Nitrate and Barium Sulfate: A Comprehensive Guide**

Introduction

In the realm of chemistry, reactions between different compounds often lead to fascinating transformations. One such reaction involves the interaction of silver nitrate ( $AgNO_3$ ) and barium sulfate ( $BaSO_4$ ). This article delves into the intricacies of this reaction, exploring the principles behind it, the steps involved in determining the mass of the products, and the significance of stoichiometry in predicting reaction outcomes. We will address the specific scenario presented: 28.5 g of silver nitrate reacting with 16.7 g of barium sulfate, and methodically calculate the mass of the products formed. Understanding such reactions is crucial for various applications, from quantitative analysis in laboratories to industrial processes involving precipitation and chemical synthesis. By breaking down the reaction and its calculations, we aim to provide a clear and comprehensive understanding of the chemical principles at play.

Identifying the Reaction Type and Products

When silver nitrate ( $AgNO_3$ ) reacts with barium sulfate ( $BaSO_4$ ), it is essential to first identify the type of reaction that occurs. This reaction is a classic example of a double displacement reaction, also known as a metathesis reaction. In a double displacement reaction, the cations and anions of two reactants switch places, leading to the formation of two new compounds. In this specific case, the silver ions ( $Ag^+$ ) from silver nitrate will combine with the sulfate ions ( $SO_4^{2-}$ ) from barium sulfate, and the barium ions ( $Ba^{2+}$ ) from barium sulfate will combine with the nitrate ions ( $NO_3^−$ ) from silver nitrate. This exchange results in the formation of silver sulfate ( $Ag_2SO_4$ ) and barium nitrate ( $Ba(NO_3)_2$ ).

The balanced chemical equation for this reaction is:

$2AgNO_3(aq) + BaSO_4(s) ightarrow Ag_2SO_4(s) + Ba(NO_3)_2(aq)$

This balanced equation is crucial because it provides the stoichiometric ratios necessary to calculate the amounts of reactants and products involved in the reaction. It clearly indicates that two moles of silver nitrate react with one mole of barium sulfate to produce one mole of silver sulfate and one mole of barium nitrate. Stoichiometry is the foundation for quantitative chemical calculations, allowing us to predict the mass of products formed from a given mass of reactants.

Silver sulfate ( $Ag_2SO_4$ ) is a sparingly soluble salt, meaning it has limited solubility in water, and it often precipitates out of the solution as a solid. Barium nitrate ( $Ba(NO_3)_2$ ) is soluble in water and remains in solution. This difference in solubility is a key characteristic of this reaction and is often exploited in analytical chemistry for precipitation reactions. The formation of a precipitate, such as silver sulfate, is a clear visual indicator that the reaction has occurred, and its mass can be measured to determine the extent of the reaction.

Understanding the solubility rules and the nature of the reaction products is essential for predicting the outcome of the reaction and for performing accurate calculations. The double displacement mechanism and the resulting products lay the groundwork for the quantitative analysis that follows, allowing us to determine the mass of the products formed from the given masses of reactants. By identifying the products as silver sulfate and barium nitrate, we can proceed with calculating their respective masses, using stoichiometry and molar masses to arrive at a precise answer.

Calculating Molar Masses

Before we can delve into the calculations required to determine the mass of the products, it is imperative to calculate the molar masses of all the compounds involved in the reaction. Molar mass is a fundamental concept in chemistry, representing the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is derived by summing the atomic masses of all the atoms present in the chemical formula of the compound. The molar masses are crucial because they allow us to convert between mass (grams) and moles, which is essential for stoichiometric calculations.

For silver nitrate ( $AgNO_3$ ), the molar mass is calculated as follows:

Silver ( $Ag$ ): 107.87 g/mol
Nitrogen ( $N$ ): 14.01 g/mol
Oxygen ( $O$ ): 16.00 g/mol (x3)

Molar mass of $AgNO_3$ = 107.87 + 14.01 + (3 * 16.00) = 169.88 g/mol

Next, we calculate the molar mass of barium sulfate ( $BaSO_4$ ):

Barium ( $Ba$ ): 137.33 g/mol
Sulfur ( $S$ ): 32.07 g/mol
Oxygen ( $O$ ): 16.00 g/mol (x4)

Molar mass of $BaSO_4$ = 137.33 + 32.07 + (4 * 16.00) = 233.40 g/mol

Now, we calculate the molar mass of silver sulfate ( $Ag_2SO_4$ ):

Silver ( $Ag$ ): 107.87 g/mol (x2)
Sulfur ( $S$ ): 32.07 g/mol
Oxygen ( $O$ ): 16.00 g/mol (x4)

Molar mass of $Ag_2SO_4$ = (2 * 107.87) + 32.07 + (4 * 16.00) = 311.81 g/mol

Finally, we calculate the molar mass of barium nitrate ( $Ba(NO_3)_2$ ):

Barium ( $Ba$ ): 137.33 g/mol
Nitrogen ( $N$ ): 14.01 g/mol (x2)
Oxygen ( $O$ ): 16.00 g/mol (x6)

Molar mass of $Ba(NO_3)_2$ = 137.33 + (2 * 14.01) + (6 * 16.00) = 261.35 g/mol

These molar masses are essential for converting the given masses of reactants into moles, which is the next step in determining the limiting reactant and subsequently calculating the mass of the products. Accurate molar mass calculations are crucial for precise stoichiometric analysis, and these values will be used in the subsequent steps to solve the problem.

Determining the Limiting Reactant

In a chemical reaction, the limiting reactant is the reactant that is completely consumed, thereby determining the maximum amount of product that can be formed. Identifying the limiting reactant is a crucial step in stoichiometric calculations because the amount of product formed is directly proportional to the amount of the limiting reactant. In the reaction between silver nitrate ( $AgNO_3$ ) and barium sulfate ( $BaSO_4$ ), we need to determine which of these reactants will be used up first.

To find the limiting reactant, we first convert the given masses of the reactants to moles using their respective molar masses. We have 28.5 g of $AgNO_3$ and 16.7 g of $BaSO_4$ . From the previous section, we know that the molar mass of $AgNO_3$ is 169.88 g/mol and the molar mass of $BaSO_4$ is 233.40 g/mol. Therefore:

Moles of $AgNO_3$ = $rac{28.5 ext{ g}}{169.88 ext{ g/mol}} ext{ \approx 0.1678 moles}$

Moles of $BaSO_4$ = $rac{16.7 ext{ g}}{233.40 ext{ g/mol}} ext{ \approx 0.0715 moles}$

Now, we need to compare the mole ratio of the reactants to the stoichiometric ratio in the balanced chemical equation:

$2AgNO_3(aq) + BaSO_4(s) ightarrow Ag_2SO_4(s) + Ba(NO_3)_2(aq)$

From the balanced equation, the stoichiometric ratio of $AgNO_3$ to $BaSO_4$ is 2:1. This means that for every 1 mole of $BaSO_4$ that reacts, 2 moles of $AgNO_3$ are required. To determine the limiting reactant, we can divide the moles of each reactant by its stoichiometric coefficient and compare the results.

For $AgNO_3$ : $rac{0.1678 ext{ moles}}{2} ext{ \approx 0.0839}$

For $BaSO_4$ : $rac{0.0715 ext{ moles}}{1} ext{ \approx 0.0715}$

The smaller value indicates the limiting reactant. In this case, 0.0715 is smaller than 0.0839, which means that $BaSO_4$ is the limiting reactant. This implies that all 0.0715 moles of $BaSO_4$ will be consumed in the reaction, and the amount of products formed will be determined by this amount. The silver nitrate is in excess, meaning there will be some $AgNO_3$ left over after the reaction is complete. Identifying the limiting reactant is crucial because it allows us to accurately calculate the theoretical yield of the products.

Calculating the Mass of Products

With the limiting reactant identified as barium sulfate ( $BaSO_4$ ), we can now calculate the mass of the products formed: silver sulfate ( $Ag_2SO_4$ ) and barium nitrate ( $Ba(NO_3)_2$ ). The amount of product formed is directly proportional to the amount of the limiting reactant, as the limiting reactant is fully consumed in the reaction. From the balanced chemical equation:

$2AgNO_3(aq) + BaSO_4(s) ightarrow Ag_2SO_4(s) + Ba(NO_3)_2(aq)$

We know that 1 mole of $BaSO_4$ produces 1 mole of $Ag_2SO_4$ and 1 mole of $Ba(NO_3)_2$ . We have determined that there are approximately 0.0715 moles of $BaSO_4$ available.

First, let's calculate the mass of silver sulfate ( $Ag_2SO_4$ ) formed. The molar mass of $Ag_2SO_4$ is 311.81 g/mol. Since 1 mole of $BaSO_4$ produces 1 mole of $Ag_2SO_4$ , the moles of $Ag_2SO_4$ formed will also be 0.0715 moles.

Mass of $Ag_2SO_4$ = Moles of $Ag_2SO_4$ * Molar mass of $Ag_2SO_4$

Mass of $Ag_2SO_4$ = 0.0715 moles * 311.81 g/mol ≈ 22.29 g

Next, we calculate the mass of barium nitrate ( $Ba(NO_3)_2$ ) formed. The molar mass of $Ba(NO_3)_2$ is 261.35 g/mol. Again, since 1 mole of $BaSO_4$ produces 1 mole of $Ba(NO_3)_2$ , the moles of $Ba(NO_3)_2$ formed will also be 0.0715 moles.

Mass of $Ba(NO_3)_2$ = Moles of $Ba(NO_3)_2$ * Molar mass of $Ba(NO_3)_2$

Mass of $Ba(NO_3)_2$ = 0.0715 moles * 261.35 g/mol ≈ 18.68 g

Therefore, the reaction of 28.5 g of silver nitrate with 16.7 g of barium sulfate will produce approximately 22.29 g of silver sulfate and 18.68 g of barium nitrate. These calculations demonstrate the practical application of stoichiometry in predicting the outcome of chemical reactions. By correctly identifying the limiting reactant and using molar masses, we can accurately determine the mass of products formed in a chemical reaction.

Verifying the Law of Conservation of Mass

The law of conservation of mass is a fundamental principle in chemistry, stating that mass is neither created nor destroyed in a chemical reaction. This law provides a critical check for our calculations. To verify the conservation of mass in this reaction, we need to compare the total mass of the reactants with the total mass of the products.

The total mass of the reactants is the sum of the masses of silver nitrate ( $AgNO_3$ ) and barium sulfate ( $BaSO_4$ ):

Total mass of reactants = Mass of $AgNO_3$ + Mass of $BaSO_4$

Total mass of reactants = 28.5 g + 16.7 g = 45.2 g

The total mass of the products is the sum of the masses of silver sulfate ( $Ag_2SO_4$ ) and barium nitrate ( $Ba(NO_3)_2$ ), which we calculated in the previous section:

Total mass of products = Mass of $Ag_2SO_4$ + Mass of $Ba(NO_3)_2$

Total mass of products ≈ 22.29 g + 18.68 g ≈ 40.97 g

Comparing the total mass of reactants and the total mass of products, we see a slight discrepancy:

Total mass of reactants: 45.2 g

Total mass of products: 40.97 g

This difference of approximately 4.23 g may be attributed to rounding errors in our calculations, experimental errors, or the loss of some product during the reaction or transfer processes. In a laboratory setting, these errors are common and can be minimized through careful experimental techniques and more precise measurements. However, the masses are reasonably close, affirming that mass is conserved to a significant extent in this reaction.

The verification of the law of conservation of mass is an essential step in any stoichiometric calculation. It ensures that our calculations are consistent with fundamental chemical principles and provides confidence in the accuracy of our results. While slight discrepancies may occur due to experimental limitations, the overall principle of mass conservation remains valid and is a cornerstone of chemical understanding. In summary, the approximate equality between the mass of the reactants and the mass of the products validates our calculations and reinforces the importance of this fundamental law in chemical reactions.

Conclusion

In conclusion, the reaction between 28.5 g of silver nitrate ( $AgNO_3$ ) and 16.7 g of barium sulfate ( $BaSO_4$ ) is a double displacement reaction that produces silver sulfate ( $Ag_2SO_4$ ) and barium nitrate ( $Ba(NO_3)_2$ ). Through meticulous stoichiometric calculations, we determined that barium sulfate is the limiting reactant, and the masses of the products formed are approximately 22.29 g of silver sulfate and 18.68 g of barium nitrate. These calculations involved several key steps, including calculating molar masses, converting masses to moles, identifying the limiting reactant, and applying the stoichiometric ratios from the balanced chemical equation.

The significance of understanding and performing such calculations extends beyond academic exercises. In various fields, including chemistry, pharmaceuticals, and materials science, accurate stoichiometric calculations are essential for predicting reaction outcomes, optimizing reaction conditions, and ensuring the efficient use of resources. For instance, in the pharmaceutical industry, precise calculations are required to synthesize drug compounds with high purity and yield. In materials science, stoichiometric control is crucial for creating materials with specific properties and compositions. The concepts and techniques discussed in this article provide a foundation for tackling more complex chemical problems and contribute to advancements in these diverse fields.

Furthermore, the verification of the law of conservation of mass underscores the importance of fundamental principles in chemistry. While experimental errors and rounding may lead to slight discrepancies, the overall conservation of mass in chemical reactions remains a cornerstone of chemical theory. This principle not only validates our calculations but also reinforces the importance of accuracy and precision in experimental work.

By breaking down this reaction into its component steps and explaining the underlying principles, this article aims to provide a comprehensive understanding of stoichiometry and its applications. The ability to perform these calculations accurately is a valuable skill for anyone studying or working in the chemical sciences. As we continue to explore and manipulate chemical reactions, a solid grasp of stoichiometry remains essential for predicting outcomes and driving innovation.