Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 4 - Sections 4.1-4.9 - Exercises - Problems by Topic - Page 187: 34b


$22.92g\ KBr$

Work Step by Step

$15.39g\ Br_{2}\times\frac{1mol\ Br_{2}}{159.81g\ Br_{2}}\times\frac{2mol\ KBr}{1mol\ Br_{2}}\times\frac{119.00g\ KBr}{1mol\ KBr}= 22.92g\ KBr$ We divide the mass in grams of the reactant by its molar mass to get the number of moles of reactant. we then multiply this by the ratio obtained from the balanced chemical equation with the reactant on the bottom and the product on the top. We now have the number of moles of product. We multiply this by the molar mass of the product to get our result.
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