Explanation Balanced chemical equation for the reaction given is written as follows. H 2 A + 2 XOH → X 2 A + 2 H 2 O Number of moles of reactants: Molar mass of H 2 A is given as 168 g ⋅ mol − 1 and mass of H 2 A is given as 1 .20 g . Molar mass of XOH is given as 125 g ⋅ mol − 1 and mass of XOH is given as 1 .00 g . Amount of reactant is calculated as follows. n ( H 2 A ) = 1.20 g 168 g ⋅ mol − 1 = 0.0071 mol n ( XOH ) = 1.00 g 125 g ⋅ mol − 1 = 0.008 mol From the balanced chemical equation, it is found that one mole of H 2 A reacts with two mole of XOH to form one mole of X 2 A . The number of moles of X 2 A that can be formed from each reactant is calculated as follows. n ( X 2 A ) = 0.0071 mol H 2 A × 1 mol X 2 A 1 mol H 2 A = 0.0071 mol X 2 A n ( X 2 A ) = 0.008 mol XOH × 1 mol X 2 A 2 mol XOH = 0

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

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ISBN: 9781319403959

Author: ATKINS

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Chapter F, Problem M.18E

Interpretation Introduction

Interpretation:

Percentage yield of the reaction has to be calculated.

Concept Introduction:

The yield that can be obtained to the maximum from the given quantity of the reactant is known as theoretical yield. This does not consider the side reactions or impurities in the product formation. Percentage yield is the yield of the product that is formed expressed in terms of percentage. This can be given by the equation given below;

Percentage yield=Actual yieldTheoretical yield×100

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Chapter F, Problem M.17E

Chapter F, Problem M.19E

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Chapter F Solutions

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

Ch. F - Prob. A.1AST Ch. F - Prob. A.1BST Ch. F - Prob. A.2AST Ch. F - Prob. A.2BST Ch. F - Prob. A.3AST Ch. F - Prob. A.3BST Ch. F - Prob. A.4AST Ch. F - Prob. A.4BST Ch. F - Prob. A.5AST Ch. F - Prob. A.5BST

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Solution Summary: The author explains that the theoretical yield is the yield of the product that is formed expressed in percentage. The balanced chemical equation for the reaction is written as follows.

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