Question

In a \(1-\) L beaker, \(203 \mathrm{mL}\) of \(0.307 M\) ammonium chromate was mixed with \(137 \mathrm{mL}\) 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 reaction is \(2\,(NH_4)_2CrO_4 + 3\,Cr(NO_2)_3 \rightarrow 12\,NH_4NO_2 + Cr_2(CrO_4)_3\). The mass of chromium(III) chromate isolated with an 88.0% percent yield is approximately \(6.875\,\mathrm{g}\).

Step-by-step solution

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the reaction between ammonium chromate and chromium(III) nitrite, which produce ammonium nitrite and chromium(III) chromate. The chemical formulas of the reactants and products are: - Ammonium chromate: (NH4)2CrO4 - Chromium(III) nitrite: Cr(NO2)3 - Ammonium nitrite: NH4NO2 - Chromium(III) chromate: Cr2(CrO4)3 Now, we write the chemical equation and balance it: \(2\,(NH_4)_2CrO_4 + 3\,Cr(NO_2)_3 \rightarrow 12\,NH_4NO_2 + Cr_2(CrO_4)_3\)

Calculate the mass of chromium(III) chromate isolated

We have the balanced chemical equation and the percent yield. Now, we can use stoichiometry to find the mass of chromium(III) chromate isolated. First, we need to find the limiting reactant. To do this, we use the moles of each reactant and the stoichiometry of the balanced chemical equation: - Moles of ammonium chromate = volume × concentration = \(203\,\mathrm{mL} \times 0.307\,\mathrm{M} = 0.0623\,\mathrm{mol}\) - Moles of chromium(III) nitrite = volume × concentration = \(137\,\mathrm{mL} \times 0.269\,\mathrm{M} = 0.0368\,\mathrm{mol}\) Comparing moles of reactants given their stoichiometry: \(0.0623\,\mathrm{mol}\,(NH_4)_2CrO_4 \times \frac{3\,\mathrm{mol}\,Cr(NO_2)_3}{2\,\mathrm{mol}\,(NH_4)_2CrO_4} \approx 0.0934\,\mathrm{mol}\,Cr(NO_2)_3\) Since \(0.0368\,\mathrm{mol} < 0.0934\,\mathrm{mol}\), chromium(III) nitrite is the limiting reactant. Next, we determine the moles of chromium(III) chromate that can be produced: \(0.0368\,\mathrm{mol}\,Cr(NO_2)_3 \times \frac{1\,\mathrm{mol}\,Cr_2(CrO_4)_3}{3\,\mathrm{mol}\,Cr(NO_2)_3} \approx 0.0123\,\mathrm{mol}\,Cr_2(CrO_4)_3\) Since the percent yield of the reaction is 88.0%, we need to calculate the actual moles of chromium(III) chromate produced: \(0.0123\,\mathrm{mol} \times 0.880 = 0.0108\,\mathrm{mol}\,Cr_2(CrO_4)_3\) Finally, we calculate the mass of chromium(III) chromate isolated: mass = moles × molar mass The molar mass of chromium(III) chromate = \(2\times 51.9961 + 3\times (51.9961 + 4\times 15.999) = 635.9748\,\mathrm{g/mol}\) mass = \(0.0108\,\mathrm{mol} \times 635.9748\,\mathrm{g/mol} \approx 6.875\,\mathrm{g}\) So, the mass of chromium(III) chromate isolated is approximately \(6.875\,\mathrm{g}\).

Key concepts

Balancing Chemical Equations

Understanding the basics of balancing chemical equations is the foundation of chemical stoichiometry. Let's take the starting point from the given exercise. We have a reaction between ammonium chromate and chromium(III) nitrite forming ammonium nitrite and chromium(III) chromate. To balance the equation, we need an equal number of each type of atom on both sides of the equation.

Through trial and error or systematic approach, coefficients are assigned to the compounds to achieve this balance. For instance, the balanced equation provided from the exercise, \(2(NH_4)_2CrO_4 + 3Cr(NO_2)_3 \rightarrow 12NH_4NO_2 + Cr_2(CrO_4)_3\), shows we need 2 moles of ammonium chromate to react with 3 moles of chromium(III) nitrite to produce 12 moles of ammonium nitrite and 1 mole of chromium(III) chromate.

This balanced equation is not only crucial for understanding the reaction's stoichiometry but also for the next steps in the problem-solving process.

Limiting Reactant Calculation

Once you've balanced the chemical equation, it’s possible to identify the limiting reactant. The limiting reactant is the compound that will be used up first in the reaction, thus determining the amount of product that can be formed. In the provided exercise, we used the moles of each reactant and the balanced equation to find that chromium(III) nitrite is the limiting reactant.

Here's a closer look: given the volumes and molarities of the reactants, convert these to moles; then compare the ratio of the reactant moles to their coefficients in the balanced equation. If you find that one reactant has fewer moles than required, it's the limiting reactant. In the exercise, chromium(III) nitrite is the one that runs out, and therefore, limits the amount of product formed.

Percent Yield

Once you know the limiting reactant, you can predict the theoretical yield—the maximum amount of product that can be obtained from the reactants. However, in real-world conditions, not every reaction goes to completion, and there are often losses due to various factors. Thus, the actual yield is usually less than the theoretical yield. Percent yield becomes a measure of a reaction's efficiency and is calculated by \( \frac{{\text{{Actual Yield}}}}{{\text{{Theoretical Yield}}}} \times 100\% \).

In the given exercise, the percent yield was given as 88%. Using this information and the theoretical yield calculated from the stoichiometry, we can determine the mass of the chromium(III) chromate that was actually produced in the experiment, not just the amount that could have been produced if the reaction were perfect. This contextualizes the reaction within real laboratory conditions, offering a more practical understanding of chemical processes.

Molar Mass

The molar mass of a compound is the weight of one mole of that substance and is particularly useful for converting between grams and moles in a chemical reaction. The molar mass is the sum of the masses of all atoms in the molecule. For chromium(III) chromate as seen in the exercise, the molar mass is a sum of the atomic masses of chromium and oxygen.

Finding the molar mass of chromium(III) chromate, a complex molecule, can be challenging, but carefully adding up the atomic masses of the constituent chromium and oxygen atoms from the periodic table will get you there. To solve the exercise we used the calculated molar mass to convert moles of chromium(III) chromate to grams, which gives us the mass of the product that was actually recovered from the reaction.