Chapter 9, Problem 66QAP

### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

Chapter
Section

### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
70 views

# Solid copper can be produced by passing gaseous ammonia over solid copper (II) oxide al high temperatures.mg src=Images/HTML_99425-9-66QAP_image001.jpg alt="" align="top"/>e other products of the reaction are nitrogen gas and water Vapor. The balanced equation for this reaction is::math> 2 NH 3 ( g ) + 3 CuO ( s ) → N 2 ( g ) + 3 Cu ( s ) + 3 H 2 O ( g ) at is the theoretical yield of solid copper that should form when 18.1 g of NH 3 , is reacted with 90.4 g of CuO ? If only 45.3 g of copper is actually collected, what is the percent yield?mg src=Images/HTML_99425-9-66QAP_image002.jpg alt="" align="top"/>

Interpretation Introduction

Interpretation:

To determine the theoretical yield and percentage yield of copper.

Concept introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

The limiting reactant in a particular reaction has due to following properties:

1. Limiting reactant completely reacted in a particular reaction.
2. Limiting reactant determines the amount of the product in mole.
Explanation

Theoretical yield is also known as calculated yield. Theoretical yield is the amount of product which is theatrically calculated by the amount of limiting agent. It is the maximum amount of product which is formed in any reaction.

Other name of actual yield is observed yield.

The percentage yield is calculated by the actual or observed experimental yield divided by the theoretical yield and multiplied by 100.

The percentage yield can be calculates by the use of following expression:

Actual yield or given yield Theoretical yield or calculated yield ×100%

Theoretical yield is the greatest amount of a product possible from a reaction, when assume that the reaction goes to 100% competition.

The balance equation for the calcium and oxalate ions is as follows:

2NH3(g)+3CuO(s)  N2(g)+3Cu(s)+3H2O(g)

Given:

Amount of NH3 = 18. 1 g

Amount of CuO = 90.4 g

Calculation:

Number of moles of NH3 and CuO are calculated as follows:

Number of moles=mass in gmolarmass=18.1 g17.031  g/mol=1.06 moles NH3Number of moles=mass in gmolarmass=90

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