Answer
17.0 g of $Al_2O_3$ can be produced.
Work Step by Step
1. Calculate the molar mass of $O_2$:
$O: 16.00g * 2= 32.00g $
$ \frac{1 mole (O_2)}{ 32.00g (O_2)}$ and $ \frac{ 32.00g (O_2)}{1 mole (O_2)}$
2. The balanced reaction is:
$4Al(s) + 3O_2(g) --\gt 2Al_2O_3$
According to the coefficients, the ratio of $O_2$ to $Al_2O_3$ is 3 to 2:
$ \frac{ 2 moles(Al_2O_3)}{ 3 moles (O_2)}$ and $ \frac{ 3 moles (O_2)}{ 2 moles(Al_2O_3)}$
3. Calculate the molar mass for $Al_2O_3$:
$Al: 26.98g * 2= 53.96g $
$O: 16.00g * 3= 48.00g $
53.96g + 48.00g = 101.96g
$ \frac{1 mole (Al_2O_3)}{ 101.96g (Al_2O_3)}$ and $ \frac{ 101.96g (Al_2O_3)}{1 mole (Al_2O_3)}$
4. Use the conversion factors to find the mass of $Al_2O_3$
$8.00g(O_2) \times \frac{1 mole(O_2)}{ 32.00g( O_2)} \times \frac{ 2 moles(Al_2O_3)}{ 3 moles (O_2)} \times \frac{ 101.96 g (Al_2O_3)}{ 1 mole (Al_2O_3)} = 17.0g (Al_2O_3)$