Answer
$6.02$ g of $O_2$ are required to react with 30.0 g of $PbS$.
Work Step by Step
1. Calculate the molar mass of $PbS$:
$Pb: 207.2g $
$S: 32.07g$
207.2g + 32.07g = 239.3g
$ \frac{1 mole (PbS)}{ 239.3g (PbS)}$ and $ \frac{ 239.3g (PbS)}{1 mole (PbS)}$
2. The balanced reaction is:
$2PbS(s) + 3O_2(g) --\gt 2PbO(s) + 2SO_2(g)$
According to the coefficients, the ratio of $PbS$ to $O_2$ is 2 to 3:
$ \frac{ 3 moles(O_2)}{ 2 moles (PbS)}$ and $ \frac{ 2 moles (PbS)}{ 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 $O_2$
$30.0g(PbS) \times \frac{1 mole(PbS)}{ 239.3g( PbS)} \times \frac{ 3 moles(O_2)}{ 2 moles (PbS)} \times \frac{ 32.00 g (O_2)}{ 1 mole (O_2)} = 6.02g (O_2)$
