## Chemistry: Atoms First (2nd Edition)

a. 137.99 g/mol b. 32 g c. $7.2 \times 10^{23}$ mol d. 480 g
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: $$3.5 \space mole \times \frac{ 137.99 \space g}{1 \space mole} = 480 \space g$$