Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 6 - Chemical Composition - Exercises - Problems - Page 198: 48


a) $116.16 g CF_{4}$ b) $34.58 g MgF_{2}$ c) $0.098 g CS_{2}$ d) $151,313.4 g SO_{3}$

Work Step by Step

Using the molar mass of the compound provided, we can simply convert from grams to moles. a) $1.32 moles CF_{4}\times\frac{88.00 g/mole}{1 mole CF_{4}} = 116.16 g CF_{4}$ b) $0.555 moles =MgF_{2}\times\frac{62.30 g/mole}{1 mole CF_{4}} = 34.58 g MgF_{2}$ c) $1.29 mmoles = 0.00129 moles CS_{2}\times\frac{76.14 g/mole}{1 mole CS_{2}} = 0.098 g CS_{2}$ d) $1.89 kmol = 1890 moles SO_{3}\times\frac{80.06 g/mole}{1 mole SO_{3}} = 151,313.4 g SO_{3}$
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