The deep blue compound \( \text{Cu(NH}_3\text{)}_4\text{SO}_4 \) is made by the reaction of copper(II) sulfate and ammonia.\[\text{CuSO}_4(\text{aq}) + 4 \, \text{NH}_3(\text{aq}) \rightarrow \text{Cu(NH}_3\text{)}_4\text{SO}_4(\text{aq})\]1. If you use 31.0 g of \( \text{CuSO}_4 \) and excess \( \text{NH}_3 \), what is the theoretical yield of \( \text{Cu(NH}_3\text{)}_4\text{SO}_4 \)? \[ \_\_\_\_\_\_\_\_\_\_ \, \text{g} \, \text{Cu(NH}_3\text{)}_4\text{SO}_4 \]2. If you isolate 27.0 g of \( \text{Cu(NH}_3\text{)}_4\text{SO}_4 \), what is the percent yield of \( \text{Cu(NH}_3\text{)}_4\text{SO}_4 \)? \[ \_\_\_\_\_\_\_\_\_\_ \, \% \]

Molar mass of CuSO4 = 159.5 g Molar mass of Cu(NH3)4SO4 = 227.5 g From the balanced chemical…

Answered: e percent yield of Cu(NH3)4SO4?

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e percent yield of Cu(NH3)4SO4?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

The deep blue compound \( \text{Cu(NH}_3\text{)}_4\text{SO}_4 \) is made by the reaction of copper(II) sulfate and ammonia.

\[
\text{CuSO}_4(\text{aq}) + 4 \, \text{NH}_3(\text{aq}) \rightarrow \text{Cu(NH}_3\text{)}_4\text{SO}_4(\text{aq})
\]

1. If you use 31.0 g of \( \text{CuSO}_4 \) and excess \( \text{NH}_3 \), what is the theoretical yield of \( \text{Cu(NH}_3\text{)}_4\text{SO}_4 \)?

\[ \_\_\_\_\_\_\_\_\_\_ \, \text{g} \, \text{Cu(NH}_3\text{)}_4\text{SO}_4 \]

2. If you isolate 27.0 g of \( \text{Cu(NH}_3\text{)}_4\text{SO}_4 \), what is the percent yield of \( \text{Cu(NH}_3\text{)}_4\text{SO}_4 \)?

\[ \_\_\_\_\_\_\_\_\_\_ \, \% \]

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