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Limiting Reactants (Section) Solutions of sulfuric acid

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 82 Chapter 3

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 82

Limiting Reactants (Section) Solutions of sulfuric acid and lead(ll) acetate react to form solid lead(ll) sulfate and a solution of acetic acid. If 5.00 g of sulfuric acid and 5.00 g of lead(ll) acetate are mixed, calculate the number of grams of sulfuric acid, lead(ll) acetate, lead(ll) sulfate, and acetic acid present in the mixture after the reaction is complete.

Step-by-Step Solution:

Step-by-step solution Step 1 of 2 The limiting reagent (or limiting reactant) in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. The balanced equation for the reaction of lead (II)acetate and sulfuric acid as follows : Pb(CH3COO)2(aq)+ H2SO4(aq)PbSO4(s)+2CH3COOH(aq) Here one mole of lead acetate reacts with one mole of sulfuric acid as follows Molar mass of lead acetate =325.29 g/mol Molar mass of sulfuric acid =98.08g/mol . So moles of 5.00g sulfuric acid =(5.00g/98.08*1 mol) =0.0510 molH2SO4 Moles of 5.00g of pb(CH3COO)2=(5.00g/325.29g*1 mol) 0.0154 mol Pb(CH3COO)3

Step 2 of 2

Chapter 3, Problem 82 is Solved
Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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Limiting Reactants (Section) Solutions of sulfuric acid