Question

In a \(1-\mathrm{L}\) beaker, \(203 \mathrm{~mL}\) of \(0.307 \mathrm{M}\) 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 \((NH_4)_2CrO_4 + Cr(NO_2)_3 \rightarrow 2 NH_4NO_2 + Cr_2(CrO_4)_3\). The limiting reactant is chromium(III) nitrite, and the mass of chromium(III) chromate isolated is 16.83 g.

Step-by-step solution

1. Write the balanced chemical equation for the reaction.

First, let's write the chemical formulas for the reaction: Ammonium chromate: \((NH_4)_2CrO_4\) Chromium(III) nitrite: \(Cr(NO_2)_3\) Ammonium nitrite: \(NH_4NO_2\) Chromium(III) chromate: \(Cr_2(CrO_4)_3\) Now we write the balanced chemical equation: \((NH_4)_2CrO_4 + Cr(NO_2)_3 \rightarrow 2 NH_4NO_2 + Cr_2(CrO_4)_3\)

2. Calculate the moles of each reactant.

Next, we will calculate the number of moles for each reactant using the volume and molarity given: Moles of ammonium chromate = volume × molarity = 203 mL × 0.307 M = 0.203 L × 0.307 mol/L = 0.0623 mol Moles of chromium(III) nitrite = volume × molarity = 137 mL × 0.269 M = 0.137 L × 0.269 mol/L = 0.0368 mol

3. Determine the limiting reactant.

Now we need to determine the limiting reactant. We will compare the mole ratios between the reactants: Mole ratio ammonium chromate / chromium(III) nitrite = 0.0623 mol / 0.0368 mol = 1.69 From the balanced chemical equation, we can see that the mole ratio for ammonium chromate to chromium(III) nitrite should be 1:1. Since our calculated mole ratio is greater than 1, the limiting reactant is chromium(III) nitrite.

4. Calculate the theoretical yield of chromium(III) chromate.

Next, we determine the theoretical yield of chromium(III) chromate using the mole ratio between the limiting reactant and the product: Moles of chromium(III) chromate = moles of limiting reactant = 0.0368 mol Now, we will calculate the mass of chromium(III) chromate using the molecular weight: Molecular weight of chromium(III) chromate = 2(51.996) + 3(51.996 + 4(15.999)) = 519.86 g/mol Theoretical yield of chromium(III) chromate = moles × molecular weight = 0.0368 mol × 519.86 g/mol = 19.13 g

5. Calculate the actual yield of chromium(III) chromate.

We are given the percent yield of the reaction, which is 88.0%. To calculate the actual yield, we will multiply the theoretical yield by the percent yield: Actual yield of chromium(III) chromate = theoretical yield × percent yield = 19.13 g × 0.88 = 16.83 g Therefore, the mass of chromium(III) chromate isolated is 16.83g.

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that gets completely used up first. It limits the extent of the reaction and determines the amount of product formed. In our example, we have two reactants: **ammonium chromate** and **chromium(III) nitrite**.
To identify the limiting reactant, we compare the moles of each reactant. Using the formula:

• Moles = Volume (L) × Molarity (mol/L),

we calculate:

• **Ammonium Chromate:** 0.0623 mol
• **Chromium(III) Nitrite:** 0.0368 mol

The balanced equation indicates a 1:1 mole ratio. Since there are fewer moles of chromium(III) nitrite, it is the limiting reactant. It runs out first, controlling the amount of chromium(III) chromate we can make.

Percent Yield

In chemistry, reactions often do not go to completion, or other factors may prevent the maximum amount of product from being formed. Percent yield tells us how efficient a reaction was. It compares the actual yield - the amount of product we obtain - to the theoretical yield - the calculated maximum amount possible. Percent yield is calculated using the formula:

• % Yield = (Actual Yield / Theoretical Yield) × 100

In this exercise, the reaction's percent yield is 88%. Therefore, if the theoretical yield of chromium(III) chromate was 19.13 grams, after accounting for the reaction's efficiency, the actual yield was 16.83 grams.

Theoretical Yield

The theoretical yield is the maximum amount of product that could be formed from the given quantities of reactants. It assumes that every molecule of reactant forms product, which is not always the case in reality.
To find the theoretical yield, first identify the limiting reactant, then use stoichiometry based on the balanced chemical equation. Finally, calculate the mass of the product produced.
In our scenario:

• The limiting reactant, chromium(III) nitrite, provides 0.0368 moles.
• The balanced equation provides a 1:1 relationship with the product, so 0.0368 moles of chromium(III) chromate should theoretically form.
• Using chromium(III) chromate's molar mass of 519.86 g/mol, the theoretical yield calculates to 19.13 grams.

Balanced Chemical Equation

Writing a correct balanced chemical equation is crucial. It shows the stoichiometry of a reaction, the correct proportions of reactants and products.
In the given problem:

• Ammonium Chromate is \((NH_4)_2CrO_4\).
• Chromium(III) Nitrite is \(Cr(NO_2)_3\).
• Ammonium Nitrite is \(NH_4NO_2\).
• Chromium(III) Chromate is \(Cr_2(CrO_4)_3\).

From these, we construct the balanced equation:\[(NH_4)_2CrO_4 + Cr(NO_2)_3 \rightarrow 2 NH_4NO_2 + Cr_2(CrO_4)_3\]This equation shows the reactants and products in their proper mole ratios, essential for any further calculations like determining limiting reactants, theoretical yield, and percent yield.