   # 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 chrornium(III) chromate was isolated? ### Chemistry

9th Edition
Steven S. Zumdahl
Publisher: Cengage Learning
ISBN: 9781133611097

#### Solutions

Chapter
Section ### Chemistry

9th Edition
Steven S. Zumdahl
Publisher: Cengage Learning
ISBN: 9781133611097
Chapter 4, Problem 141IP
Textbook Problem
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## 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 chrornium(III) chromate was isolated?

Interpretation Introduction

Interpretation: The mass of chromium (III) chromate has to be isolated.

Concept Introduction:

The amount of products can be defined as the ratio of the actual yield to theoretical yield. The amount of products can be calculated by using the formula,

Amountofproduct=actualyieldtheoreticalyield

### Explanation of Solution

Explanation

Record the given data

Volume of Ammonium chromate = 203mL

Molarity of ammonium chromate = 0.307M

Volume of chromium (III) nitrate= 137mL

Molarity of chromium (III) nitrate= 0.269M

The volumes and molarities of ammonium chromate and chromium (III) nitrate are recorded as shown above.

To calculate the moles of chromium (III) chromate

Molar mass of chromium (III) chromate= 452.00g/mol

Moles of chromium (III) chromate (with ammonium chromate as reagent) =0.2030.307mol(NH4)2CrO4L×1molCr2(CrO4)33mol(NH4)2CrO4×452.00gCr2(CrO4)3molCr2(CrO4)3

= 9.39gCr2(CrO4)3

Moles of chromium (III) chromate (with chromium (III) nitrate as reagent)

=0.1370.269molCr(NO2)3L×1molCr2(CrO4)32molCr(NO2)3×452.00gCr2(CrO4)3molCr2(CrO4)3

= 8

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