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: 36b


$44.0g\ Na_{2}CO_{3}$

Work Step by Step

Divide the mass of the second reactant by its molar mass to find the moles of the second reactant. Multiply this by the mole ratio from the balanced chemical equation to get the moles of the first reactant. Finally, multiply the moles of the first reactant by the molar mass of the the first reactant to get the mass of the first reactant. $55.8g\ CuCl_{2}\times\frac{1mol\ CuCl_{2}}{134.452g\ CuCl_{2}}\times\frac{1mol\ CuCl_{2}}{1mol\ Na_{2}CO_{3}}\times\frac{105.988g\ Na_{2}CO_{3}}{1mol\ Na_{2}CO_{3}}=44.0g\ Na_{2}CO_{3}$
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