   # Ammonia reacts with O 2 to form either NO( g ) or NO 2 ( g ) according to these unbalanced equations: NH 3 ( g ) + O 2 ( g ) → NO ( g ) + H 2 O ( g ) NH 3 ( g ) + O 2 ( g ) → NO 2 ( g ) + H 2 O ( g ) In a certain experiment 2.00 moles of NH 3 (g) and 10.00 moles of O 2 ( g ) are contained in a closed flask. After the reaction is complete, 6.75 moles ofO 2 ( g ) remains. Calculate the number of moles of NO(g) in the product mixture: ( Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.) ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 5, Problem 177CP
Textbook Problem
521 views

## Ammonia reacts with O2 to form either NO(g) or NO2(g) according to these unbalanced equations: NH 3 ( g ) + O 2 ( g ) → NO ( g ) + H 2 O ( g ) NH 3 ( g ) + O 2 ( g ) → NO 2 ( g ) + H 2 O ( g ) In a certain experiment 2.00 moles of NH3(g) and 10.00 moles of O2(g) are contained in a closed flask. After the reaction is complete, 6.75 moles ofO2(g) remains. Calculate the number of moles of NO(g) in the product mixture: (Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)

Interpretation Introduction

Interpretation: The number of moles of NO(g) in the product mixture is calculated.

Concept introduction: The number of moles in an atom is defined as the given mass divided by the molecular mass. One mole is equal to the Avogadro’s number and each element consists of different molar mass that depends on the weight of its atom.

To determine: The number of moles of NO(g) in the product mixture.

### Explanation of Solution

Explanation

Given

The balanced chemical reactions are,

4NH3+5O24NO+6H2O (1)

4NH3+7O24NO2+6H2O (2)

Number of moles of NH3(g) is 2.00 moles .

Number of moles of O2(g) is 10.00 moles .

The remaining number of moles of oxygen after the reaction is 6.75 moles .

Therefore,

Number of moles of oxygen reacted is 10.00 moles6.75 moles=3.25 moles

The number of moles of oxygen consumed for the reaction (1) is assumed to be x and the number of moles of oxygen consumed for the reaction (2) is assumed to be y .

So, the total number of moles of oxygen consumed is.

x+y=3.25 (3)

From the reaction (1), 5 moles of O2 and 4 moles of NH3 is consumed.

For x moles of O2 , the number of moles of NH3 consumed is,

(x5)4=(45)x .

From the reaction (2), 7 moles of O2 and 4 moles of NH3 is consumed.

For y moles of O2 , the number of moles of NH3 consumed is,

(y7)4=(47)y .

Therefore, the total number of moles of NH3 consumed is,

(45)x+(47)y=2

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