Answer
$3.79$ g of $N_2$ can be produced when 6.50 g of $O_2$ reacts.
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:
$4NH_3 + 3O_2 --\gt 2N_2 + 6H_2O$
According to the coefficients, the ratio of $O_2$ to $N_2$ is 3 to 2:
$ \frac{ 2 moles(N_2)}{ 3 moles (O_2)}$ and $ \frac{ 3 moles (O_2)}{ 2 moles(N_2)}$
3. Calculate the molar mass for $N_2$:
$N: 14.01g * 2= 28.02g $
$ \frac{1 mole (N_2)}{ 28.02g (N_2)}$ and $ \frac{ 28.02g (N_2)}{1 mole (N_2)}$
4. Use the conversion factors to find the mass of $N_2$
$6.50g(O_2) \times \frac{1 mole(O_2)}{ 32.00 g( O_2)} \times \frac{ 2 moles(N_2)}{ 3 moles (O_2)} \times \frac{ 28.02 g (N_2)}{ 1 mole (N_2)} = 3.79g (N_2)$
