Answer
There is a total of 0.245 mole in 25.0 g of $Al_2O_3$
Work Step by Step
1. Determine the molar mass of this compound ($Al_2O_3$), and setup the conversion factors:
Molar mass :
$Al: 26.98g * 2= 53.96g $
$O: 16.00g * 3= 48.00g $
53.96g + 48.00g = 101.96g
$ \frac{1 \space mole \space (Al_2O_3)}{ 101.96 \space g \space (Al_2O_3)}$ and $ \frac{ 101.96 \space g \space (Al_2O_3)}{1 \space mole \space (Al_2O_3)}$
2. Calculate the number of moles $(Al_2O_3)$
$ 25.0g \times \frac{1 mole}{ 101.96g} = 0.245 mole$
