Problem 79 - Chemistry: The Central Science

Question

What mass of \(\mathrm{KCl}\) is needed to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) solution?

Step-by-Step Solution

Answer

The mass of KCl required to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) solution is approximately \(0.224 \mathrm{g}\).

Step 1: List the given information and identify the target variable

We have the following information: - Volume of AgNO₃ solution = 15.0 mL - Concentration of AgNO₃ solution = 0.200 M - KCl needed to precipitate the silver ions Our target variable is the mass of KCl.

Step 2: Convert volume to liters

Convert the volume of the AgNO₃ solution from mL to L: \(15.0 \mathrm{~mL} = 15.0 \times 10^{-3} \mathrm{~L}\)

Step 3: Calculate moles of AgNO₃

Using the given concentration, calculate the moles of AgNO₃: Moles of AgNO₃ = Concentration × Volume = (0.200 M) × (15.0 × 10⁻³ L) Moles of AgNO₃ = 3.0 × 10⁻³ mol

Step 4: Determine stoichiometric ratio

According to the balanced chemical equation, the stoichiometric ratio between AgNO₃ and KCl is 1:1.

Step 5: Calculate moles of KCl

Since the stoichiometric ratio is 1:1, the moles of KCl required will be equal to the moles of AgNO₃. Moles of KCl = Moles of AgNO₃ = 3.0 × 10⁻³ mol

Step 6: Calculate mass of KCl

Using the molar mass of KCl (39.10 g/mol for K and 35.45 g/mol for Cl), find the mass of KCl required: Mass of KCl = Moles of KCl × Molar mass of KCl Mass of KCl = (3.0 × 10⁻³ mol) × (39.10 g/mol + 35.45 g/mol) Mass of KCl = (3.0 × 10⁻³ mol) × (74.55 g/mol) Mass of KCl ≈ 0.224 g The mass of KCl required to precipitate the silver ions from 15.0 mL of 0.200 M AgNO₃ solution is approximately 0.224 g.