Question

Calculate the mass of AgCl formed, and the concentration of silver ion remaining in solution, when \(10.0 \mathrm{~g}\) of solid \(\mathrm{AgNO}_{3}\) is added to \(50 . \mathrm{mL}\) of \(1.0 \times 10^{-2} \mathrm{M} \mathrm{NaCl}\) solution. Assume there is no volume change upon addition of the solid.

Short answer

The mass of AgCl formed is approximately 0.0717 g, and the concentration of silver ions remaining in the solution is approximately 1.17 M.

Step-by-step solution

Write the balanced chemical equation for the reaction

The reaction between silver nitrate (AgNO₃) and sodium chloride (NaCl) can be written as: \( AgNO_3 + NaCl \rightarrow AgCl \downarrow + NaNO_3 \) Note that AgCl will precipitate out of the solution, as indicated by the downward arrow.

Convert the mass of AgNO₃ to moles

To calculate the mass of AgCl formed in the reaction, first, we need to convert the mass of AgNO₃ to moles. The molar mass of AgNO₃ is approximately 169.87 g/mol. Use the equation: Moles of AgNO₃ = (Mass of AgNO₃) / (Molar Mass of AgNO₃) Moles of AgNO₃ = \( \frac{10.0 ~g}{169.87 ~g/mol} \approx 0.0589 ~mol \)

Determine the moles of NaCl in the solution

The initial concentration of NaCl solution is given as 1.0 × 10⁻² M in 50 mL of solution. First, convert the volume to liters: Volume (L) = \( \frac{50 ~mL}{1000} = 0.050 ~L \) Then, calculate the moles of NaCl: Moles of NaCl = (Concentration of NaCl) × (Volume of NaCl solution) Moles of NaCl = \( (1.0 \times 10^{-2} M) \times (0.050 ~L) = 5.0 \times 10^{-4} ~mol \)

Determine the limiting reactant

Comparing the moles of both reactants, we can see which one is the limiting reactant: Moles of AgNO₃ : Moles of NaCl = \( \frac{0.0589 ~mol}{5.0 \times 10^{-4} ~mol} \) The ratio is greater than 1, which means there are more moles of AgNO₃ compared to NaCl. So, the limiting reactant is NaCl, as it will be completely consumed in the reaction.

Calculate the moles of AgCl formed

Since the molar ratio of AgNO₃ to AgCl is 1:1 in the chemical equation, the moles of AgCl formed are equal to the moles of the limiting reactant, NaCl: Moles of AgCl formed = Moles of NaCl = 5.0 × 10⁻⁴ mol

Calculate the mass of AgCl formed

Now that we know the moles of AgCl, we can calculate its mass. The molar mass of AgCl is approximately 143.32 g/mol. Use the equation: Mass of AgCl = (Moles of AgCl) × (Molar Mass of AgCl) Mass of AgCl = \( (5.0 \times 10^{-4} ~mol) \times (143.32 ~g/mol) \approx 0.0717 ~g \)

Determine the moles of AgNO₃ remaining

To find out the moles of AgNO₃ remaining in the solution, subtract the moles of the limiting reactant (NaCl) from the initial moles of AgNO₃: Moles of AgNO₃ remaining = Moles of AgNO₃ - Moles of NaCl Moles of AgNO₃ remaining = 0.0589 mol - 5.0 × 10⁻⁴ mol = 0.0584 mol

Calculate the concentration of silver ions remaining in the solution

Finally, to calculate the concentration of silver ions remaining in the 50 mL solution, use the equation: \( Concentration_{Ag^+} = \frac{Moles_{AgNO₃ \, remaining}}{Volume_{solution}} \) \( Concentration_{Ag^+} = \frac{0.0584 ~mol}{0.050 ~L} \approx 1.17 ~M \) The concentration of silver ions remaining in the solution is about 1.17 M. So, the mass of AgCl formed is approximately 0.0717 g, and the concentration of silver ions remaining in the solution is approximately 1.17 M.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the law of conservation of mass, which states that matter cannot be created or destroyed in an isolated system. In simple terms, stoichiometry helps us understand the quantitative relationships in a chemical equation.

For example, consider the balanced chemical equation given in the exercise:

• AgNO₃ + NaCl → AgCl + NaNO₃

This equation tells us that one mole of silver nitrate ( AgNO₃ ) reacts with one mole of sodium chloride ( NaCl ) to produce one mole of silver chloride ( AgCl ) and one mole of sodium nitrate ( NaNO₃ ).

• We can use this relationship to determine how much of any substance is needed or produced.
• This is where stoichiometry steps in to help us calculate the needed values for the reaction.

In the exercise, stoichiometry allows us to relate moles of AgNO₃ to moles of AgCl using the mole-to-mole ratios from the balanced equation. Being comfortable with these calculations is crucial for solving many chemistry problems.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is consumed first, which stops the reaction from continuing because no more product can be formed. Identifying the limiting reactant is essential because it determines the maximum amount of product that can be generated.

In the given exercise, we compare the amount of moles of each reactant:

• AgNO₃: 0.0589 mol
• NaCl: 5.0 × 10⁻⁴ mol

The chemical equation for such a reaction shows a 1:1 ratio between AgNO₃ and NaCl , indicating both react with equal efficiency. Since NaCl is present in a smaller amount relative to AgNO₃ , it becomes the limiting reactant.

Knowing the limiting reactant allows us to calculate the amount of AgCl formed. As we see in the step-by-step exercise, 5.0 × 10⁻⁴ moles of NaCl yield an equal amount of AgCl due to the 1:1 stoichiometric ratio. Recognizing and calculating the limiting reactant is critical for predicting the outcome of chemical reactions.

Molarity

Molarity (M) is a unit of concentration, specifically moles per liter, that describes the amount of a solute in a given volume of solution. Understanding and calculating molarity is key when preparing solutions and performing reactions in chemistry.

For instance, in the exercise, the molarity of the initial NaCl solution is given as 1.0 imes 10^{-2} M. This tells us that in every liter of this solution, there are 1.0 imes 10^{-2} moles of NaCl. With this value known, we multiply by the volume of solution (in liters) to find the total moles of NaCl present:

• Moles of NaCl = Molarity × Volume = \(1.0 \times 10^{-2}\) M × 0.050 L = 5.0 × 10⁻⁴ mol

Molarity serves as a convenient way to describe solution concentration, allowing calculations with ease. This is particularly important when reactions are carried out in solutions, and knowing molarity helps ensure the correct amount of reactants is used.

Moles

The mole is a fundamental unit in chemistry that measures the quantity of a substance. One mole of any substance contains Avogadro's number of entities, 6.022 imes 10^{23}, thus linking macroscopic and microscopic worlds. Moles allow chemists to count particles by weighing them.

To find how many moles are present, we use the formula: Moles = Mass /Molar Mass. In the step-by-step exercise solution, the mass of AgNO₃ is converted to moles:

• Moles of AgNO₃ = \( \frac{10.0 ~g}{169.87 ~g/mol}\)
• Moles of AgNO₃ ≈ 0.0589 ~mol

The usage of moles simplifies calculations. For example, instead of counting individual molecules of AgNO₃, which is impractical, we consider the total in moles when reacting with NaCl. Understanding how to convert between mass, moles, and number of particles is a critical skill in chemistry.