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?

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: \( 3(NH_4)_2CrO_4 + 2Cr(NO_2)_3 \rightarrow 6NH_4NO_2 + Cr_2(CrO_4)_3 \)

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.

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.

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.