Answer
51.4 g of $SO_2$
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:
$2H_2S(g) + 3O_2(g) --\gt 2SO_2(g) + 2H_2O(g)$
According to the coefficients, the ratio of $O_2$ to $SO_2$ is 3 to 2:
$ \frac{ 2 moles(SO_2)}{ 3 moles (O_2)}$ and $ \frac{ 3 moles (O_2)}{ 2 moles(SO_2)}$
3. Calculate the molar mass for $SO_2$:
$S: 32.07g $
$O: 16.00g * 2= 32.00g $
32.07g + 32.00g = 64.07g
$ \frac{1 mole (SO_2)}{ 64.07g (SO_2)}$ and $ \frac{ 64.07g (SO_2)}{1 mole (SO_2)}$
4. Use the conversion factors to find the mass of $SO_2$
$38.5g(O_2) \times \frac{1 mole(O_2)}{ 32.00g( O_2)} \times \frac{ 2 moles(SO_2)}{ 3 moles (O_2)} \times \frac{ 64.07 g (SO_2)}{ 1 mole (SO_2)} = 51.4g (SO_2)$
