Answer
19.2 g of $O_2$ are needed to react with 13.6 g of $NH_3$
Work Step by Step
1. Calculate the molar mass of $NH_3$:
$N: 14.01g $
$H: 1.008g * 3= 3.024g $
14.01g + 3.024g = 17.03g
$ \frac{1 mole (NH_3)}{ 17.03g (NH_3)}$ and $ \frac{ 17.03g (NH_3)}{1 mole (NH_3)}$
2. The balanced reaction is:
$4NH_3 + 3O_2 --\gt 2N_2 + 6H_2O$
According to the coefficients, the ratio of $NH_3$ to $O_2$ is 4 to 3:
$ \frac{ 3 moles(O_2)}{ 4 moles (NH_3)}$ and $ \frac{ 4 moles (NH_3)}{ 3 moles(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 $O_2$
$13.6g(NH_3) \times \frac{1 mole(NH_3)}{ 17.03g( NH_3)} \times \frac{ 3 moles(O_2)}{ 4 moles (NH_3)} \times \frac{ 32.00 g (O_2)}{ 1 mole (O_2)} = 19.2g (O_2)$
