   # The reaction of 750. g each of NH 3 and O 2 was found to produce 562 g of NO (see pages 177-179). 4 NH 3 (g) + 5 O 2 (g) → 4 NO(g) + 6 H 2 O(l) (a) What mass of water is produced by this reaction? (b) What mass of O 2 is required to consume 750. g of NH 3 ? ### 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 84GQ
Textbook Problem
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## The reaction of 750. g each of NH3 and O2 was found to produce 562 g of NO (see pages 177-179).4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l)(a) What mass of water is produced by this reaction?(b) What mass of O2 is required to consume 750. g of NH3?

(a)

Interpretation Introduction

Interpretation:

The mass of water produced and the mass of oxygen require to react with 750.0ofNH3 in the given reaction should be determined.

Concept introduction:

Numberofmole=GivenmassofthesubstanceMolarmass

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• 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.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• 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.

### Explanation of Solution

The balanced chemical reaction equation for the given reaction is,

4NH3(g)+5O2(g)4NO(g)+6H2O(l)

The amount (moles) of NH3andO2  available can be calculated by using the equation

Numberofmole=GivenmassofthesubstanceMolarmass

Amount of   NH3 = 750.0gNH3×1molNH317gNH3 = 44.1molNH3

Amount of O2 =750gO2×1molO232gO2 =23

(b)

Interpretation Introduction

Interpretation:

The mass of O2 required to consume 750.0g of NH3 has to be determined.

Concept introduction:

Numberofmole=GivenmassofthesubstanceMolarmass

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• 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.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• 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.

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