Question

What mass of solid AgBr is produced when \(100.0 \mathrm{mL}\) of 0.150 M AgNO_is added to \(20.0 \mathrm{mL}\) of \(1.00 \mathrm{M}\) NaBr?

Short answer

The mass of solid AgBr produced when 100.0 mL of 0.150 M AgNO₃ is added to 20.0 mL of 1.00 M NaBr is approximately 2.82 g.

Step-by-step solution

1. Write the balanced chemical equation for the reaction

The reaction between silver nitrate and sodium bromide can be written as: AgNO₃(aq) + NaBr(aq) → AgBr(s) + NaNO₃(aq) Both reactants are in aqueous form, and the reaction produces a solid silver bromide (AgBr) precipitate and an aqueous sodium nitrate (NaNO₃) as the by-product.

2. Convert volumes to moles

We are given volume and concentration, so we need to convert them to moles (n) using the formula: n = C × V For AgNO₃: Volume (V₁) = 100.0 mL = 0.100 L (convert mL to L by dividing by 1000) Concentration (C₁) = 0.150 M n₁ = C₁ × V₁ = (0.150 mol/L) × (0.100 L) = 0.0150 mol of AgNO₃ For NaBr: Volume (V₂) = 20.0 mL = 0.020 L Concentration (C₂) = 1.00 M n₂ = C₂ × V₂ = (1.00 mol/L) × (0.020 L) = 0.0200 mol of NaBr

3. Determine the limiting reactant

Based on the stoichiometry of the balanced chemical equation, 1 mole of AgNO₃ reacts with 1 mole of NaBr. Therefore, we can compare the mole ratio of the reactants to determine which one is the limiting reactant: Mole ratio= \( \frac{n_{1}}{n_{2}} \) = \( \frac{0.0150}{0.0200} \) = 0.75 Since the mole ratio is less than 1, it means AgNO₃ is the limiting reactant, and the reaction will stop when all AgNO₃ is used up.

4. Calculate the moles of AgBr produced

Now, we need to find the moles of AgBr produced, which will be equal to the moles of the limiting reactant, AgNO₃: n₃ = n₁ = 0.0150 mol of AgBr

5. Convert moles of AgBr to mass

Finally, we need to convert the moles of AgBr to mass using the molecular weight of the compound: Molecular weight of AgBr = 107.87 g/mol (Ag) + 79.90 g/mol (Br) = 187.77 g/mol Mass of AgBr = n₃ × molecular weight of AgBr = (0.0150 mol) × (187.77 g/mol) = 2.8165 g The mass of solid AgBr produced when 100.0 mL of 0.150 M AgNO₃ is added to 20.0 mL of 1.00 M NaBr is approximately 2.82 g.

Key concepts

Stoichiometry

Stoichiometry is akin to a recipe for chemistry. It is a quantitative examination of the relationships between the amounts of reactants and products in a chemical reaction. Integral to mastering stoichiometry is understanding the principles of a balanced chemical equation, where the number of atoms for each element is equal on both sides of the reaction. This balance ensures the law of conservation of mass is upheld.

When solving stoichiometry problems, like determining the amount of AgBr solid produced from a reaction, you begin with a balanced equation and use it as a guide to relate the quantities of one reactant or product to another. This relationship is commonly referred to as the mole ratio, derived from coefficients in the balanced equation. The step-by-step solution provided for the AgBr problem illustrates how stoichiometry helps you calculate the amounts of reagents and products involved in the reaction. It's useful to approach these problems methodically by first establishing the mole ratio and proceeding to calculate the quantity of interest, often in moles or mass.

Balanced Chemical Equation

A balanced chemical equation provides a snapshot of a chemical reaction where the number of atoms for each element is the same on both the reactant and product sides. In the context of the AgNO₃ and NaBr reaction, the balanced equation is essential as it provides the stoichiometric coefficients that tell us the mole ratios needed to completely react the reagents without any excess.

This balancing act isn't just academic; it reflects an essential law of chemistry: the law of conservation of mass. In a chemical reaction, atoms are neither created nor destroyed, only rearranged. The balanced equation for the formation of AgBr from AgNO₃ and NaBr, represented as AgNO₃(aq) + NaBr(aq) → AgBr(s) + NaNO₃(aq), shows a one-to-one ratio of reactants to products. Grasping this concept is pivotal for accurately determining the limiting reactant and, subsequently, predicting the quantity of product formed.

Mole Conversion

The use of the mole, the fundamental unit of amount in chemistry, allows scientists and students to count by weighing. Mole conversion is the process of converting between mass, volume, and number of particles using the mole concept. It plays a crucial role in stoichiometry, uniting disparate concepts under a common metric.

In the given exercise, mole conversion is used in multiple steps to determine the mass of AgBr produced. First, concentrations (in molarity) and volumes are converted to moles using the formula \( n = C \times V \). This provides the basis for determining the limiting reactant. Once the limiting reactant is established, mole conversions allow us to translate the number of moles into a measurable mass, making use of the molar mass of AgBr. By mastering mole conversions, one can navigate between the macroscopic world we measure and the atomic world we conceptualize.