Answer
$54.0$ g of $H_2O$ are formed from the reaction of 34.0 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 $H_2O$ is 4 to 6:
$ \frac{ 6 moles(H_2O)}{ 4 moles (NH_3)}$ and $ \frac{ 4 moles (NH_3)}{ 6 moles(H_2O)}$
3. Calculate the molar mass for $H_2O$:
$H: 1.008g * 2= 2.016g $
$O: 16.00g$
2.016g + 16g = 18.02g
$ \frac{1 mole (H_2O)}{ 18.02g (H_2O)}$ and $ \frac{ 18.02g (H_2O)}{1 mole (H_2O)}$
4. Use the conversion factors to find the mass of $H_2O$
$34.0g(NH_3) \times \frac{1 mole(NH_3)}{ 17.03g( NH_3)} \times \frac{ 6 moles(H_2O)}{ 4 moles (NH_3)} \times \frac{ 18.02 g (H_2O)}{ 1 mole (H_2O)} = 54.0g (H_2O)$
