Question

In a 1-L beaker, \(203 \mathrm{~mL}\) of \(0.307 \mathrm{M}\) ammonium chromate was mixed with \(137 \mathrm{~mL}\) of \(0.269 \mathrm{M}\) chromium(III) nitrite to produce ammonium nitrite and chromium(III) chromate. Write the balanced chemical 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 reaction is: \( 3(NH_4)_2CrO_4 + 2Cr(NO_2)_3 \rightarrow 6NH_4NO_2 + Cr_2(CrO_4)_3 \) Chromium(III) nitrite is the limiting reactant. The mass of chromium(III) chromate produced in this reaction, considering the 88% percent yield, is 3.178 grams.

Step-by-step solution

Write the balanced chemical equation

The reactants are ammonium chromate (\( (NH_4)_2CrO_4 \)) and chromium(III) nitrite (\( Cr(NO_2)_3 \)). The products are ammonium nitrite (\( NH_4NO_2 \)) and chromium(III) chromate (\( Cr_2(CrO_4)_3 \)). The balanced chemical equation is: \( 3(NH_4)_2CrO_4 + 2Cr(NO_2)_3 \rightarrow 6NH_4NO_2 + Cr_2(CrO_4)_3 \)

Find the stoichiometric ratio

The balanced chemical equation gives us the stoichiometric ratio between the reactants and the product: 3 moles ammonium chromate: 2 moles chromium(III) nitrite: 1 mole chromium(III) chromate.

Determine the limiting reactant

To determine the limiting reactant, we need to calculate the moles of each reactant that are present. For ammonium chromate: moles = volume × concentration = (203 mL) × (0.307 mol/L) = 62.321 × 10⁻³ mol For chromium(III) nitrite: moles = volume × concentration = (137 mL) × (0.269 mol/L) = 36.833 × 10⁻³ mol Now, we need to find the amount of each reactant that would be required for the reaction to go to completion. We will do this using the stoichiometric ratio from Step 2. For ammonium chromate: (62.321 × 10⁻³ mol) / 3 = 20.774 × 10⁻³ mol For chromium(III) nitrite: (36.833 × 10⁻³ mol) / 2 = 18.416 × 10⁻³ mol The smaller amount is chromium(III) nitrite, so it is the limiting reactant.

Calculate the mass of chromium(III) chromate produced

Since chromium(III) nitrite is the limiting reactant, the amount of chromium(III) chromate produced will be based on the moles of chromium(III) nitrite. Using the stoichiometric ratio (1 mole of chromium(III) chromate per 2 moles of chromium(III) nitrite), we can calculate the moles of chromium(III) chromate produced: moles of chromium(III) chromate = (18.416 × 10⁻³ mol) × (1/2) = 9.208 × 10⁻³ mol We are given the percent yield of the reaction (88%). To determine the actual amount of chromium(III) chromate produced, we will multiply the moles of chromium(III) chromate by the percent yield: Actual moles of chromium(III) chromate = (9.208 × 10⁻³ mol) × 0.88 = 8.103 × 10⁻³ mol Now we can find the mass of chromium(III) chromate produced using the molar mass: mass = moles × molar mass = (8.103 × 10⁻³ mol) × (392.2 g/mol) = 3.178 g So, the mass of chromium(III) chromate produced in this reaction is 3.178 grams.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In simpler terms, it tells you how much of each substance you need to start with and how much you will get at the end of the reaction. This is achieved by using the balanced chemical equation, which provides the exact proportions of different elements needed for the reaction.
In the given exercise, the balanced chemical equation is:- \(3(NH_4)_2CrO_4 + 2Cr(NO_2)_3 \rightarrow 6NH_4NO_2 + Cr_2(CrO_4)_3\)
This equation helps us understand the stoichiometric relationships, where 3 moles of ammonium chromate react with 2 moles of chromium(III) nitrite to produce 1 mole of chromium(III) chromate along with ammonium nitrite. By converting the given volumes of the solutions into moles, using their respective molarity values, we can use stoichiometry to find out how much of each compound was necessary for this reaction.

Limiting Reactant

The limiting reactant is the reactant in a chemical reaction that is entirely consumed when the chemical reaction is complete. Due to its complete consumption, it limits the amount of product formed. To figure out which reactant is limiting in a reaction, you compare the mole ratio of the reactants used to the ratio in the balanced equation.
In this exercise, both ammonium chromate and chromium(III) nitrite are used as reactants. Calculating their moles:

• Ammonium chromate: \(203 \, \text{mL} \times 0.307 \, \text{M} = 62.321 \times 10^{-3} \, \text{mol}\)
• Chromium(III) nitrite: \(137 \, \text{mL} \times 0.269 \, \text{M} = 36.833 \times 10^{-3} \, \text{mol}\)

Comparing to the stoichiometric ratio:

• \(\frac{62.321}{3} = 20.774 \times 10^{-3}\, \text{mol}\) of ammonium chromate
• \(\frac{36.833}{2} = 18.416 \times 10^{-3}\, \text{mol}\) of chromium(III) nitrite

The limiting reactant is the one with the smaller quotient, which is chromium(III) nitrite in this case. Thus, the amount of chromium(III) chromate formed will depend on chromium(III) nitrite.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction. It compares the actual yield of a product (what you obtained in practice) to the theoretical yield (what you should have obtained if everything went perfectly). The percent yield is calculated as follows:\[\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\]In this exercise, after determining chromium(III) nitrite as the limiting reactant, the theoretical yield of chromium(III) chromate is calculated based on this reactant using stoichiometric relationships. However, the reaction only achieved an 88% yield:
The theoretical yield based on molar calculations was \(9.208 \times 10^{-3} \, \text{mol}\) of chromium(III) chromate. Applying the percent yield:

• Actual moles of chromium(III) chromate: \(9.208 \times 10^{-3} \, \text{mol} \times 0.88 = 8.103 \times 10^{-3} \, \text{mol}\)
• Mass of chromium(III) chromate: \(8.103 \times 10^{-3} \, \text{mol} \times 392.2 \, \text{g/mol} = 3.178\, \text{g}\)

This calculated mass shows the actual yield of chromium(III) chromate given the percent yield, emphasizing the necessity of understanding both theoretical calculations and practical limitations in chemistry.