Answer
$17.4$ g of $SO_2$ can be produced when 65.0 g of $PbS$ reacts.
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 $SO_2$ is 2 to 2:
$ \frac{ 2 moles(SO_2)}{ 2 moles (PbS)}$ and $ \frac{ 2 moles (PbS)}{ 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 $SO_2$
$65.0g(PbS) \times \frac{1 mole(PbS)}{ 239.3g( PbS)} \times \frac{ 2 moles(SO_2)}{ 2 moles (PbS)} \times \frac{ 64.07 g (SO_2)}{ 1 mole (SO_2)} = 17.4g (SO_2)$
