Answer
94.9 g of $Al_2O_3$ are produced when 50.2 g of Al reacts.
Work Step by Step
1. Calculate the molar mass of $Al$:
$Al: 26.98g$
$ \frac{1 mole (Al)}{ 26.98g (Al)}$ and $ \frac{ 26.98g (Al)}{1 mole (Al)}$
2. The balanced reaction is:
$4Al(s) + 3O_2(g) --\gt 2Al_2O_3$
According to the coefficients, the ratio of $Al$ to $Al_2O_3$ is 4 to 2:
$ \frac{ 2 moles(Al_2O_3)}{ 4 moles (Al)}$ and $ \frac{ 4 moles (Al)}{ 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 $Al_2O_3$
$50.2g(Al) \times \frac{1 mole(Al)}{ 26.98g( Al)} \times \frac{ 2 moles(Al_2O_3)}{ 4 moles (Al)} \times \frac{ 101.96 g (Al_2O_3)}{ 1 mole (Al_2O_3)} = 94.9g (Al_2O_3)$
