Problem 6 - Chemistry and Chemical Reactivity

Question

The formation of water-insoluble silver chloride is useful in the analysis of chloride-containing substances. Consider the following unbalanced equation: $$\mathrm{BaCl}_{2}(\mathrm{aq})+\mathrm{AgNO}_{3}(\mathrm{aq}) \rightarrow \mathrm{AgCl}(\mathrm{s})+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})$$ (a) Write the balanced equation. (b) What mass of \(\mathrm{AgNO}_{3}\), in grams, is required for complete reaction with 0.156 g of \(\mathrm{BaCl}_{2} ?\) What mass of AgCl is produced?

Step-by-Step Solution

Answer

(a) \( \mathrm{BaCl}_{2} + 2 \mathrm{AgNO}_{3} \rightarrow 2 \mathrm{AgCl} + \mathrm{Ba}(\mathrm{NO}_{3})_{2} \); (b) 0.2545 g of \( \mathrm{AgNO}_{3} \), 0.2147 g of \( \mathrm{AgCl} \).

Step 1: Write the unbalanced equation

The unbalanced equation for the reaction is given as: \( \mathrm{BaCl}_{2} (\mathrm{aq}) + \mathrm{AgNO}_{3} (\mathrm{aq}) \rightarrow \mathrm{AgCl} (\mathrm{s}) + \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} (\mathrm{aq}) \).

Step 2: Balance the chemical equation

To balance the equation, we need to ensure that there are equal numbers of each type of atom on both sides. The balanced chemical equation is:\[ \mathrm{BaCl}_{2} (\mathrm{aq}) + 2 \mathrm{AgNO}_{3} (\mathrm{aq}) \rightarrow 2 \mathrm{AgCl} (\mathrm{s}) + \mathrm{Ba}(\mathrm{NO}_{3})_{2} (\mathrm{aq}) \]

Step 3: Calculate the molar mass of BaCl2

The molar mass of \( \mathrm{BaCl}_{2} \) is calculated by adding the atomic masses: \( 137.33 \) g/mol for Ba + \( 2 \times 35.45 \) g/mol for Cl = \( 208.23 \) g/mol.

Step 4: Find moles of BaCl2

Use the mass of \( \mathrm{BaCl}_{2} \) given (0.156 g) to find moles: \( \frac{0.156 \text{ g}}{208.23 \text{ g/mol}} = 0.000749 \text{ mol} \).

Step 5: Use stoichiometry to find moles of AgNO3

According to the balanced equation, 1 mole of \( \mathrm{BaCl}_{2} \) reacts with 2 moles of \( \mathrm{AgNO}_{3} \). Thus, \( 0.000749 \text{ mol} \times 2 = 0.001498 \text{ mol} \) of \( \mathrm{AgNO}_{3} \) is required.

Step 6: Calculate mass of AgNO3 required

The molar mass of \( \mathrm{AgNO}_{3} \) is calculated as follows: \( 107.87 \) g/mol for Ag + \( 14.01 \) g/mol for N + \( 3 \times 16.00 \) g/mol for O = \( 169.87 \) g/mol.Thus, \( 0.001498 \text{ mol} \times 169.87 \text{ g/mol} = 0.2545 \text{ g} \) of \( \mathrm{AgNO}_{3} \) is needed.

Step 7: Calculate the mass of AgCl produced

The molar mass of \( \mathrm{AgCl} \) is 107.87 g/mol for Ag + 35.45 g/mol for Cl = 143.32 g/mol.Since 0.000749 mol of \( \mathrm{BaCl}_{2} \) produces 0.001498 mol of \( \mathrm{AgCl} \) (1:2 ratio), the mass of \( \mathrm{AgCl} \) produced is \( 0.001498 \text{ mol} \times 143.32 \text{ g/mol} = 0.2147 \text{ g} \).

Key Concepts

Stoichiometry

In the context of chemical reactions, stoichiometry is like a recipe that tells you the necessary amounts of ingredients to get the desired product. Imagine baking a cake; if the recipe calls for 2 eggs, 1 cup of sugar, and you're missing something, the resulting cake won't turn out correctly. Similarly, in a chemical reaction, stoichiometry ensures the correct "amounts" of reactants are used.
To understand stoichiometry, we look at balanced chemical equations. These equations express the conservation of matter principle, ensuring that the number of atoms for each element is the same on the reactant side and the product side.

For instance, in the reaction between \(\mathrm{BaCl}_{2}(\mathrm{aq})\) and \(\mathrm{AgNO}_{3}(\mathrm{aq})\), the balanced equation is:

• \(\mathrm{BaCl}_{2}(\mathrm{aq}) + 2 \mathrm{AgNO}_{3}(\mathrm{aq}) \rightarrow 2 \mathrm{AgCl}(\mathrm{s}) + \mathrm{Ba}(\mathrm{NO}_{3})_{2}(\mathrm{aq}) \)

Attention to coefficients is key! The numbers before the formulas indicate the molar ratio, here indicating that 1 mole of \(\mathrm{BaCl}_{2}\) reacts with 2 moles of \(\mathrm{AgNO}_{3}\). Understanding this ratio is crucial because it helps in calculating how much of each reactant is needed or how much product will be formed.
This concept translates directly into how we approach real-world chemical problems, ensuring reactions occur as expected.

Molar Mass Calculation

Calculating molar mass is a fundamental skill in chemistry that enables you to convert between mass and moles efficiently. Molar mass is the mass of one mole of a given substance (usually in grams per mole). Knowing the molar mass helps us in stoichiometry significantly.
To perform a molar mass calculation, sum up the atomic masses of all atoms in a molecule:

1. **Calculate for \(\mathrm{BaCl}_{2}\):** - **Barium (Ba):** \(137.33 \text{ g/mol}\) - **Chlorine (Cl):** 2 atoms of \(35.45 \text{ g/mol}\) - - **Total:** \(137.33 + 2 \times 35.45 = 208.23 \text{ g/mol}\)
2. **Calculate for \(\mathrm{AgNO}_{3}\):** - **Silver (Ag):** \(107.87 \text{ g/mol}\) - **Nitrogen (N):** \(14.01 \text{ g/mol}\) - **Oxygen (O):** 3 atoms of \(16.00 \text{ g/mol}\) - - **Total:** \(107.87 + 14.01 + 3 \times 16.00 = 169.87 \text{ g/mol}\)
These calculations aid not only in solving stoichiometry problems, such as finding how much reactant is needed or how much product can be produced, but they also help to predict and measure yield in chemical reactions.

Precipitation Reaction

A precipitation reaction occurs when two soluble reactants combine to form an insoluble product, called a precipitate. Let's take a closer look at the formation of the solid product. It's like mixing two liquids and seeing a solid start forming at the bottom of your container!
In the exercise involving \(\mathrm{BaCl}_{2}\) and \(\mathrm{AgNO}_{3}\), the precipitation reaction forms \(\mathrm{AgCl}(\mathrm{s})\) as a solid product:

• \(\mathrm{BaCl}_{2}(\mathrm{aq}) + 2 \mathrm{AgNO}_{3}(\mathrm{aq}) \rightarrow 2 \mathrm{AgCl}(\mathrm{s}) + \mathrm{Ba}(\mathrm{NO}_{3})_{2}(\mathrm{aq})\)

When \(\mathrm{Ag}^{+}\) ions from \(\mathrm{AgNO}_{3}\) come into contact with \(\mathrm{Cl}^{-}\) ions from \(\mathrm{BaCl}_{2}\), they form the insoluble silver chloride (\(\mathrm{AgCl}\)). This specific reaction is important in laboratory settings, especially in techniques like gravimetric analysis, where the formation and isolation of a precipitate are used to quantify the amount of a particular ion in a solution.
Recognizing the outcomes of precipitation reactions allows chemists to predict when a solid will form, assisting in designing experiments and understanding reaction pathways.