Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 5 - Exercises - Page 240d: 64

Answer

a. 137.99 g/mol b. 32 g c. $7.2 \times 10^{23}$ mol d. 480 g

Work Step by Step

a. $ Al_2O_3(H_2O)_2 $ : ( 26.98 $\times$ 2 )+ ( 16.00 $\times$ 5 )+ ( 1.008 $\times$ 4 )= 137.99 g/mol b. - Each $ Al_2O_3(H_2O)_2 $ has 2 Al atoms, thus:$$ 0.58 \space mole \space Al_2O_3(H_2O)_2 \times \frac{ 2 \space moles \ Al }{1 \space mole \space Al_2O_3(H_2O)_2 } = 1.2 \space moles \space Al $$ $ Al $ : 26.98 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$ 1.2 \space mole \times \frac{ 26.98 \space g}{1 \space mole} = 32 \space g$$ c. $$ 1.2 \space mol \space Al \times \frac{6.02 \times 10^{23} \space atoms \ Al }{1 \space mole \space Al } = 7.2 \times 10^{23} \space moles \space Al $$ d. $$ 2.1 \times 10^{24} \space Al_2O_3(H_2O)_2 \times \frac{1 \space mol}{6.02 \times 10^{23} \space Al_2O_3(H_2O)_2 } = 3.5 \space mol \space Al_2O_3(H_2O)_2 $$ - Using the molar mass as a conversion factor, find the mass in g: $$ 3.5 \space mole \times \frac{ 137.99 \space g}{1 \space mole} = 480 \space g$$
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