Answer
$6.26$ g of $O_2$ are needed for the reaction.
Work Step by Step
1. Calculate the molar mass of $Na$:
$Na: 22.99g$
$ \frac{1 mole (Na)}{ 22.99g (Na)}$ and $ \frac{ 22.99g (Na)}{1 mole (Na)}$
2. The balanced reaction is:
$4Na + O_2 --\gt 2Na_2O$
According to the coefficients, the ratio of $Na$ to $O_2$ is 4 to 1:
$ \frac{ 1 mole(O_2)}{ 4 moles (Na)}$ and $ \frac{ 4 moles (Na)}{ 1 mole(O_2)}$
3. Calculate the molar mass for $O_2$:
$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)}$
4. Use the conversion factors to find the mass of $O_2$
$18.0g(Na) \times \frac{1 mole(Na)}{ 22.99g( Na)} \times \frac{ 1 mole(O_2)}{ 4 moles (Na)} \times \frac{ 32 g (O_2)}{ 1 mole (O_2)} = 6.26g (O_2)$
