Explanation A balanced chemical equation has equal number of atoms in both reactant and product side Part (a): In order to balance the equation we have to put 3 before S i and 2 before N 2 . Therefore, the balanced equation is S i 3 N 4 ( s ) → 3 S i ( s ) + 2 N 2 ( g ) Here, the reactant breaks down to form silicon and nitrogen. Therefore, this is a decomposition reaction. Part (b): In order to balance the equation, we have to put 3 before M g . Therefore, the balanced equation is 3 M g ( s ) + N 2 ( g ) → M g 3 N 2 ( s ) Here, magnesium and nitrogen combine to form a new compound. Therefore, this is a combination reaction. Part (c): In order to balance the number of A l , H , P and O atoms on both sides of the equation, we have to put 2 before A l and 2 before H 3 P O 4 on the reactant side and 3 before H 2 and 2 before A l P O 4 on the product side. Therefore, the balanced equation is 2 A l ( s ) + 2 H 3 P O 4 ( a q ) → 3 H 2 ( g ) + 2 A l P O 4 ( a q ) . Here, A l replaces hydrogen from H 3 P O 4 to form A l P O 4 . Therefore, this is a single replacement reaction. Part (d): In order to balance the number of carbon, hydrogen and oxygen atoms, we have to put 4 before O 2 , 3 before C O 2 and 2 before H 2 O . Therefore, the balanced equation is C 3 H 4 ( g ) + 4 O 2 ( g ) → Δ 3 C O 2 ( g ) + 2 H 2 O ( g )

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)

13th Edition

ISBN: 9780134562254

Author: Karen C Timberlake

Publisher: PEARSON

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Question

Chapter 7, Problem 7.100APP

Interpretation Introduction

To determine:Balanced chemical equation and type of reaction

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Chapter 7, Problem 7.99APP

Chapter 7, Problem 7.101APP

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Chapter 7 Solutions

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)

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