Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 3 - Mass Relationships in Chemical Reactions - Worked Example - Page 84: 8


(a) 15.0g (b) 4 tablets (c) $5.18 \times 10^{22}$ molecules.

Work Step by Step

(a) As we can see in the previous worked examples, the molar mass for glucose is equal to 180.0 g/mol. $0.0833 mol \times \frac{180.0g}{1mol} = 15.0$ g. (b) $15.0$ g $\times \frac{1 tablet}{3.75g } = 4$ $tablets$. (c) $0.0833 mol \times \frac{6.022 \times 10^{23} molecules}{1 mol} = 5.18 \times 10^{22}$ molecules.
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