   # Ammonia gas can be prepared by the following reaction: CaO(s) + 2 NH 4 Cl(s) → 2 NH 3 (g) + H 2 O(g) + CaCl 2 (s) If 112 g of CaO and 224 g of NH 4 Cl are mixed, the theoretical yield of NH 3 is 68.0 g (Study Question 12). If only 16.3 g of NH 3 is actually obtained, what is its percent yield? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 4, Problem 20PS
Textbook Problem
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## Ammonia gas can be prepared by the following reaction:CaO(s) + 2 NH4Cl(s) → 2 NH3(g) + H2O(g) + CaCl2(s)If 112 g of CaO and 224 g of NH4Cl are mixed, the theoretical yield of NH3 is 68.0 g (Study Question 12). If only 16.3 g of NH3 is actually obtained, what is its percent yield?

Interpretation Introduction

Interpretation:

Mass percentage yield of  NH3 produced in the given reaction should be determined.

Concept introduction:

• Limiting reagent: limiting reagent is a reactant, which consumes completely in the chemical reaction. The quantity of the product depends on this limiting reagent, it can be determined with the help of balanced chemical equation for the reaction.
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.

### Explanation of Solution

Balanced chemical equation for the given reaction is,

CaO(s)+2NH4Cl(s)2NH3(g)+2H2O(g)+CaCl2(s)

The mass of NH3 produced can be determined by considering the amount of substance in both the reactant CaO and NH4Cl

Therefore the mass of NH3 produced is,

112gCaO×1molCaO56.0774gCaO×2molNH31molCaO×17.031gNH31molNH3=68.0286gNH3224gNH4Cl×1molNH4Cl53

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