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Chapter 4: Problem 157

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

Expert verified
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

01

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.
02

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
03

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.
04

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
05

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
06

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.

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Most popular questions from this chapter

In the spectroscopic analysis of many substances, a series of standard solutions of known concentration are measured to generate a calibration curve. How would you prepare standard solutions containing 10.0, 25.0, 50.0, 75.0, and 100. ppm of copper from a commercially produced 1000.0-ppm solution? Assume each solution has a final volume of 100.0 mL. (See Exercise 135 for definitions.)

A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.400 M HCl. The solution was then treated with an excess of aqueous chromium(III) nitrate, resulting in formation of 2.06 g of precipitate. Determine the concentration of the NaOH solution

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A \(2.20-\) g sample of an unknown acid (empirical formula \(=\) $\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3} )\( is dissolved in 1.0 \)\mathrm{L}$ of water. A titration required 25.0 mL of 0.500 M NaOH to react completely with all the acid present. Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula of the unknown acid?
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