Question

In a 1-L beaker, 203 mL of 0.307 M ammonium chromate was mixed with 137 mL of 0.269 M chromium(III) nitrite to produce ammonium nitrite and chromium(III) chromate. Write the balanced chemical equation for the reaction occurring here. If the percent yield of the reaction was 88.0%, what mass of chromium(III) chromate was isolated?

Short answer

The balanced chemical equation for the given reaction is (NH4)2CrO4(aq) + Cr(NO2)3(aq) -> 2NH4NO2(aq) + Cr2O3(s). The mass of chromium(III) chromate isolated, based on an 88.0% percent yield, is 4.93 g.

Step-by-step solution

Write the balanced chemical equation for the reaction

The balanced chemical equation for the given reaction is: (NH4)2CrO4(aq) + Cr(NO2)3(aq) -> 2NH4NO2(aq) + Cr2O3(s) Ammonium chromate reacts with chromium(III) nitrite to produce ammonium nitrite and chromium(III) chromate.

Calculate the moles of each reactant

Using the volume (in L) and concentration (M) of each reactant, we can calculate the moles of each reactant: Moles of ammonium chromate = Volume × Concentration Moles of ammonium chromate = 0.203 L × 0.307 mol/L Moles of ammonium chromate = 0.0623 mol Moles of chromium(III) nitrite = Volume × Concentration Moles of chromium(III) nitrite = 0.137 L × 0.269 mol/L Moles of chromium(III) nitrite = 0.0368 mol

Determine the limiting reactant

To determine the limiting reactant, we will use the mole ratio from the balanced chemical equation: Mole ratio: (NH4)2CrO4 : Cr(NO2)3 = 1:1 Comparing the moles of each reactant, we can see that ammonium chromate is the limiting reactant since it has a lower number of moles (0.0623) compared to chromium(III) nitrite (0.0368) in the ratio 1:1.

Calculate the theoretical yield of chromium(III) chromate

Using the stoichiometry from the balanced chemical equation and the moles of the limiting reactant, we can calculate the theoretical yield of chromium(III) chromate: Mole ratio: (NH4)2CrO4 : Cr2O3 = 1:1 Theoretical yield of Cr2O3 = Moles of limiting reactant × Mole ratio Theoretical yield of Cr2O3 = 0.0368 mol × 1/1 Theoretical yield of Cr2O3 = 0.0368 mol

Calculate the actual yield of chromium(III) chromate

Using the given percent yield, we can calculate the actual yield of chromium(III) chromate: Actual yield = Theoretical yield × Percent yield Actual yield = 0.0368 mol × 88.0% Actual yield = 0.0368 mol × 0.88 Actual yield = 0.0324 mol

Convert the actual yield in moles to mass in grams

Using the molar mass of chromium(III) chromate, we can convert the actual yield to mass: Molar mass of Cr2O3 = 2 × (51.996 g/mol Cr) + 3 × (15.999 g/mol O) = 151.99 g/mol Mass of chromium(III) chromate = Actual yield × Molar mass Mass of chromium(III) chromate = 0.0324 mol × 151.99 g/mol Mass of chromium(III) chromate = 4.93 g The mass of chromium(III) chromate isolated is 4.93 g.

Key concepts

Balancing Chemical Equations

Balancing chemical equations is like solving a pretty simple puzzle where we ensure that the number of atoms for each element involved in a reaction is the same on both the reactant and product sides.
This is done to follow the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction.
For instance, in the chemical reaction between ammonium chromate ext{((NH}_4)_2 ext{CrO}_4)} and chromium(III) nitrite ext{(Cr(NO}_2)_3}, you will start by writing down the unbalanced equation.
Then, count the number of atoms for each element on both sides, adjusting coefficients as necessary.

• Ensure that all atoms for the reactants and products are equal.
• In our example, the balanced equation looks like this: ext{((NH}_4)_2 ext{CrO}_4(aq) + ext{Cr(NO}_2)_3(aq) ightarrow 2 ext{NH}_4 ext{NO}_2(aq) + ext{Cr}_2 ext{O}_3(s)}.

Balancing equations is a foundational skill in chemistry because misbalancing can lead to incorrect stoichiometry, affecting calculations like yield and limiting reactants.

Stoichiometry

Stoichiometry is a central concept in chemistry that involves calculating the quantities of reactants and products in chemical reactions. By using stoichiometry, we can convert between moles, masses, and particles, and determine the amounts needed or formed.
In the given problem:

• You calculate the moles of reactants using the formula: Moles = Volume (in liters) \( \times \) Concentration (Molarity).
• For ammonium chromate, it's \(0.203 \, L \times 0.307 \, mol/L = 0.0623 \, mol\).
• For chromium(III) nitrite, it's \(0.137 \, L \times 0.269 \, mol/L = 0.0368 \, mol\).
• This allows you to see how much of each substance you start with and helps identify the limiting reactant later.

Stoichiometry also lets us calculate the theoretical yield, which is the maximum amount of product that can be formed from the given amounts of reactants, based on the balanced chemical equation. This is vital for understanding how efficient your reaction can be.

Limiting Reactant

In a chemical reaction, the limiting reactant is the one that is completely consumed first, determining the maximum amount of product that can be formed. Any other reactant present in excess is known as the excess reactant.
To find the limiting reactant, compare the mole ratio of the reactants as shown in your balanced equation. In our case:

• The balanced equation shows a 1:1 mole ratio between ammonium chromate and chromium(III) nitrite.
• You compare the available moles of each reactant: ammonium chromate (0.0623 mol) and chromium(III) nitrite (0.0368 mol).
• Here, we find that chromium(III) nitrite has fewer moles, making it the limiting reactant as it will run out first and stop the reaction.

Identifying the limiting reactant is crucial for calculating the exact theoretical yield of the reaction and also for minimizing waste of excess reactants in practical applications.

Percent Yield

Percent yield tells us how effective a reaction was in producing the desired product. It’s calculated as the ratio of the actual yield (the amount actually produced) to the theoretical yield (the maximum possible amount) multiplied by 100.
In this exercise, after calculating the theoretical yield of chromium(III) chromate based on the limiting reactant, we use the percent yield to find out how much was actually isolated:

• The theoretical yield of chromium(III) chromate was 0.0368 mol.
• If the percent yield was 88%, then the actual yield would be \(0.0368 \, mol \times 0.88 = 0.0324 \, mol\).

We can then find the mass of chromium(III) chromate by multiplying the actual yield in moles by its molar mass. The percent yield helps chemists determine the efficiency and practicality of chemical reactions in real-world applications.