Answer
149 g of $O_2$
Work Step by Step
1. Calculate the molar mass of $H_2O$:
$H: 1.008g * 2= 2.016g $
$O: 16.00g$
2.016g + 16.00g = 18.02g
$ \frac{1 mole (H_2O)}{ 18.02g (H_2O)}$ and $ \frac{ 18.02g (H_2O)}{1 mole (H_2O)}$
2. The balanced reaction is:
$2H_2S(g) + 3O_2(g) --\gt 2SO_2(g) + 2H_2O(g)$
According to the coefficients, the ratio of $H_2O$ to $O_2$ is 2 to 3:
$ \frac{ 3 moles(O_2)}{ 2 moles (H_2O)}$ and $ \frac{ 2 moles (H_2O)}{ 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 $O_2$
$55.8g(H_2O) \times \frac{1 mole(H_2O)}{ 18.02g( H_2O)} \times \frac{ 3 moles(O_2)}{ 2 moles (H_2O)} \times \frac{ 32.00 g (O_2)}{ 1 mole (O_2)} = 149g (O_2)$