Question

Determine the mass of product that will precipitate when \(125.2 \mathrm{~mL} 0.8015 \mathrm{M} \mathrm{AgNO}_{3}\) and \(50.00 \mathrm{~mL}\) of \(0.7850 \mathrm{M} \mathrm{Na}_{2} \mathrm{Cr} \mathrm{O}_{4}\) are combined.

Short answer

13.02 grams of Ag_2CrO_4 will precipitate.

Step-by-step solution

Write the balanced chemical equation

The reaction between silver nitrate (AgNO_3) and sodium chromate (Na_2CrO_4) can be written as follows:\[ 2 \text{AgNO}_3 + \text{Na}_2\text{CrO}_4 \rightarrow \text{Ag}_2\text{CrO}_4 \downarrow + 2 \text{NaNO}_3 \]Here, Ag_2CrO_4 is the precipitate that forms.

Calculate moles of AgNO_3

Use the concentration and volume to find the moles of AgNO_3: \[ \text{moles of AgNO}_3 = \text{Molarity} \times \text{Volume} = 0.8015 \frac{\text{mol}}{\text{L}} \times 0.1252 \text{ L} = 0.1003 \text{ moles} \]

Calculate moles of Na_2CrO_4

Use the concentration and volume to find the moles of Na_2CrO_4: \[ \text{moles of Na}_2\text{CrO}_4 = \text{Molarity} \times \text{Volume} = 0.7850 \frac{\text{mol}}{\text{L}} \times 0.0500 \text{ L} = 0.03925 \text{ moles} \]

Determine the limiting reactant

From the balanced equation, 2 moles of AgNO_3 react with 1 mole of Na_2CrO_4. Calculate how much AgNO_3 is required to react with all Na_2CrO_4:\[ 0.03925 \text{ moles of Na}_2\text{CrO}_4 \times \frac{2 \text{ moles of AgNO}_3}{1 \text{ mole of Na}_2\text{CrO}_4} = 0.0785 \text{ moles of AgNO}_3 \]Since 0.1003 moles of AgNO_3 are available, AgNO_3 is in excess, and Na_2CrO_4 is the limiting reactant.

Calculate moles of Ag_2CrO_4 precipitated

According to the balanced equation, 1 mole of Na_2CrO_4 forms 1 mole of Ag_2CrO_4. Thus, the moles of Ag_2CrO_4 will be the same as the moles of Na_2CrO_4:\[ 0.03925 \text{ moles of Ag}_2\text{CrO}_4 \]

Calculate mass of Ag_2CrO_4

The molar mass of Ag_2CrO_4 is calculated as follows:\[ 2(107.87 \text{ g/mol for Ag}) + 51.996 \text{ g/mol for Cr} + 4(16.00 \text{ g/mol for O}) = 331.73 \text{ g/mol} \]Now, calculate the mass of Ag_2CrO_4 precipitated:\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.03925 \times 331.73 = 13.02 \text{ grams} \]

Key concepts

Limiting Reactant

In chemical reactions, the concept of a limiting reactant is crucial to understanding how reactions are completed. When two or more reactants are combined, the limiting reactant is the substance that is completely consumed first, stopping the reaction from proceeding further. This reactant determines the maximum amount of product that can be formed.

In the given problem, the limiting reactant is determined by comparing the mole ratio of the reactants to what is needed for the reaction to occur. Based on the balanced equation:

• 2 moles of AgNO₃ react with 1 mole of Na₂CrO₄.
• To find out how much AgNO₃ is required to react with Na₂CrO₄, we multiply the moles of Na₂CrO₄ by 2 (the stoichiometric coefficient from the equation).
• If the available moles of AgNO₃ are greater than the required moles, then Na₂CrO₄ is the limiting reactant.

In this case, despite having 0.1003 moles of AgNO₃, which is more than needed, Na₂CrO₄ becomes the limiting reactant as only 0.03925 moles of it is available.

Precipitation Reaction

Precipitation reactions are a type of chemical reaction where two soluble salts react in solution to form one or more insoluble products, known as precipitates. These reactions are important in various fields such as chemistry and environmental science.

In the solution you have, a precipitation reaction occurs between silver nitrate (AgNO₃) and sodium chromate (Na₂CrO₄), forming the insoluble salt silver chromate (Ag₂CrO₄) as the precipitate. The balanced chemical equation is: - 2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ ↓ + 2NaNO₃
The arrow pointing downward indicates that Ag₂CrO₄ will settle out of the solution as a solid, signifying that the reaction has reached completion. Recognizing when a precipitate forms is key to determining the extent and success of a chemical reaction.

Molarity

Molarity is a term used to describe the concentration of a solution. It is defined as the number of moles of solute (the substance dissolved) per liter of solution. It is usually expressed in moles per liter (mol/L).

For example, the molarity of silver nitrate (AgNO₃) in the problem is given as 0.8015 M and that of sodium chromate (Na₂CrO₄) as 0.7850 M. To find the number of moles of each solute in their respective solutions, you multiply the molarity of the solution by the volume (in liters) of the solution:

• For AgNO₃: 0.8015 M imes 0.1252 L = 0.1003 moles
• For Na₂CrO₄: 0.7850 M imes 0.0500 L = 0.03925 moles

How molarity is used in calculations helps us understand how much of each reactant there is to participate in the reaction.

Molar Mass

Molar mass is a fundamental concept in chemistry that denotes the mass of one mole of a given substance. It is expressed in grams per mole (g/mol) and helps convert between mass and moles, which are crucial for stoichiometric calculations.

Let's calculate the molar mass for silver chromate ( Ag₂CrO₄). Based on atomic masses:

• Silver (Ag) has an atomic mass of 107.87 g/mol, and there are two silver atoms: 2 × 107.87 = 215.74 g/mol.
• Chromium (Cr) has an atomic mass of 51.996 g/mol.
• Oxygen (O) has an atomic mass of 16.00 g/mol, and there are four oxygen atoms: 4 × 16.00 = 64.00 g/mol.

Adding these together gives you the molar mass of Ag₂CrO₄ as 331.73 g/mol.

Once you know the moles of Ag₂CrO₄ were calculated to be 0.03925 moles, you can find the mass of the precipitate by multiplying the molar mass by the number of moles: 0.03925 moles × 331.73 g/mol = 13.02 grams. This illustrates how molar mass is used to determine the actual mass of substances in stoichiometric problems.