# Ammonium nitrate has been used as a high explosive because it is unstable and decomposes into several gaseous substances. The rapid expansion of the gaseous substances produces the explosive force. :math&gt; NH 4 NO 3 ( s ) → N 2 ( g ) + O 2 ( g ) + H 2 O ( g ) lculate the mass of each product gas ¡f1 .25 g of ammonium nitrate reacts.

### Introductory Chemistry: A Foundati...

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

Chapter
Section

### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781337399425
Chapter 9, Problem 37QAP
Textbook Problem
35 views

## Ammonium nitrate has been used as a high explosive because it is unstable and decomposes into several gaseous substances. The rapid expansion of the gaseous substances produces the explosive force.:math> NH 4 NO 3 ( s ) → N 2 ( g ) + O 2 ( g ) + H 2 O ( g ) lculate the mass of each product gas ¡f1 .25 g of ammonium nitrate reacts.

Interpretation Introduction

Interpretation:

The mass of each product gas from1.25 g of ammonium nitrate should be calculated.

Concept Introduction:

The mass of products can be calculated from their number of moles which can be calculated from number of moles of ammonium nitrate according to balanced chemical reaction.

### Explanation of Solution

The balanced chemical reaction is as follows:

2NH4NO3(s)2N2(g)+O2(g)+4H2O(g)

According to balanced chemical reaction, 2 mol of ammonium nitrate gives 2 mol of nitrogen gas, 1 mol of oxygen gas and 4 mol of water.

First calculate the number of moles of ammonium nitrate as follows:

n=mM

Molar mass of ammonium nitrate is 80.043 g/mol. Number of moles will be:

n=1.25 g80.043 g/mol=0.0156 mol

From the balanced chemical reaction, 2 mol of ammonium nitrate gives 2 mol of nitrogen gas or 1 mol of ammonium nitrate gives 1 mol of nitrogen gas thus, 0.0156 mol of ammonium nitrate gives 0.0156 mol of nitrogen gas.

Since, molar mass of nitrogen gas is 28 g/mol thus, mass of nitrogen gas will be:

m=n×M

Putting the values,

m=(0.0156 mol)(28 g/mol)=0

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