Question

What mass of silver chloride can be prepared by the reaction of 100.0 mL of 0.20 M silver nitrate with 100.0 mL of 0.15 M calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete

Short answer

The mass of silver chloride formed is 4.30 g, and the concentrations of the ions in the solution after precipitation are: [Ca²⁺] = 0 mol/L, [Cl⁻] = 0 mol/L, [Ag⁺] = 0.025 mol/L, and [NO₃⁻] = 0.025 mol/L.

Step-by-step solution

Write down the chemical reaction

The reactants are silver nitrate (AgNO3) and calcium chloride (CaCl2), and the product is silver chloride (AgCl) in a precipitation reaction. The balanced chemical equation for this reaction is: 2 AgNO3(aq) + CaCl2(aq) -> 2 AgCl(s) + Ca(NO3)2(aq)

Calculate the moles of reactants

Using the volume and concentration of the reactants, we can find the moles of each reactant: Moles of AgNO3 = Volume × Concentration = 100.0 mL × 0.20 mol/L = 0.02 mol Moles of CaCl2 = Volume × Concentration = 100.0 mL × 0.15 mol/L = 0.015 mol

Determine the limiting reactant

From the balanced equation, we can see that the mole ratio between AgNO3 and CaCl2 is 2:1. Therefore, we need to compare the moles of AgNO3 and CaCl2 to determine the limiting reactant: Mole ratio AgNO3 : CaCl2 = 0.02 mol : 0.015 mol = 4/3 Since 4/3 is greater than 2/1, the limiting reactant is CaCl2.

Calculate the moles of products

Using the stoichiometry of the balanced equation and the moles of the limiting reactant (CaCl2), we can find the moles of the precipitate (AgCl) formed: Moles of AgCl = Moles of CaCl2 × (2 mol AgCl / 1 mol CaCl2) = 0.015 mol × 2 = 0.03 mol

Calculate the mass of silver chloride produced

Now we can calculate the mass of the silver chloride produced using the molar mass: Mass of AgCl = Moles of AgCl × Molar mass of AgCl = 0.03 mol × (107.87 g/mol + 35.45 g/mol) = 0.03 mol × 143.32 g/mol = 4.30 g

Calculate the concentrations of ions in the solution after precipitation

The precipitation reaction consumed all the CaCl2, thus limiting the amount of AgNO3 reacted. Since the mole ratio is 2:1, the excess moles of AgNO3 are: Moles of excess AgNO3 = 0.02 mol - 2 × 0.015 mol = 0.005 mol Now we can calculate the final concentration of each ion. The total volume of the solution is 200.0 mL. Concentration of Ca2+: 0 mol/L (since all CaCl2 is used up) Concentration of Cl-: 0 mol/L (since all Cl- ions react with Ag+ to form AgCl) Concentration of Ag+: Moles of excess AgNO3 / Total volume = 0.005 mol / 0.2 L = 0.025 mol/L Concentration of NO3-: Total moles of NO3- in the solution after reaction / Total volume = (2 × 0.02 mol - 2 × 0.015 mol) / 0.2 L = 0.025 mol/L So, the mass of silver chloride formed is 4.30 g, and the concentrations of the ions in the solution after precipitation are: [Ca2+] = 0 mol/L, [Cl-] = 0 mol/L, [Ag+] = 0.025 mol/L, and [NO3-] = 0.025 mol/L.

Key concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is completely consumed first, hence determining the maximum amount of product that can be formed. It's crucial because it limits the reaction's progress.

To find the limiting reactant, compare the moles of each reactant based on the balanced chemical equation. For the reaction between silver nitrate (AgNO3) and calcium chloride (CaCl2) to form silver chloride (AgCl), the equation is: \[ 2 \text{AgNO}_3(aq) + \text{CaCl}_2(aq) \rightarrow 2 \text{AgCl}(s) + \text{Ca(NO}_3)_2(aq) \]

Here, the ratio is 2:1 for AgNO3 to CaCl2. By calculating the moles of each reactant based on their volume and concentration, you can identify which one is limiting. In this problem, calcium chloride (CaCl2) is the limiting reactant because the necessary moles of AgNO3 exceed the available moles when compared using the required stoichiometric ratios.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships of reactants and products in a chemical reaction. It tells us how much of each substance is involved or produced in a reaction.

The stoichiometry of our balanced equation informs us about the ratios of reactants and products. For the reaction involving AgNO3 and CaCl2, the stoichiometry determines that each mole of CaCl2 reacts with two moles of AgNO3 to produce two moles of AgCl.

Using the limiting reactant (CaCl2), you can employ stoichiometry to calculate the moles of AgCl formed:

• Moles of AgCl = (moles of CaCl2) \( \times \) (2 moles of AgCl / 1 mole of CaCl2)
• Thus, 0.015 mol of CaCl2 produces 0.03 mol of AgCl.

This understanding helps in determining theoretical yields and actual amounts produced in a reaction.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in g/mol. It is a critical concept for converting between moles and grams.

To compute the mass of silver chloride (AgCl) formed, you first need the molar mass of AgCl, which is the sum of the atomic masses of silver (Ag) and chloride (Cl). Using atomic weights:

• Silver (Ag) has an atomic mass of 107.87 g/mol
• Chloride (Cl) has an atomic mass of 35.45 g/mol
• Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol

Hence, the mass of AgCl formed can be calculated as:

• Mass of AgCl = \( 0.03 \text{ mol} \times 143.32 \text{ g/mol} = 4.30 \text{ g} \)

Molar mass helps relate the amount of substance in moles to its mass in grams, providing a bridge to practical laboratory measurements.

Concentration Calculation

Concentration calculation involves determining the amount of a substance in a given volume and is fundamental for preparing solutions and analyzing reaction yields. In chemistry, concentrations are often expressed as molarity (M), or moles per liter.

After identifying the excess reactants and determining the remaining ions in the solution, calculate their concentrations. Here, only excess AgNO3 remains after all CaCl2 has reacted. The concentrations are computed by dividing the moles of each ion by the total volume of the solution.

For this exercise:

• Excess moles of AgNO3 = 0.005 mol
• Total solution volume = 0.2 L (100 mL + 100 mL = 200 mL)
• Concentration of Ag+ = \( \frac{0.005 \text{ mol}}{0.2 \text{ L}} = 0.025 \text{ M} \)
• Concentration of NO3- = \( \frac{0.005 \text{ mol}}{0.2 \text{ L}} = 0.025 \text{ M} \)

This accurate concentration calculation helps predict the behavior of remaining ions in the solution after the reaction.