Answer
$137$ g of $PbS$ are used to produce $128$ g of $PbO$.
Work Step by Step
1. Calculate the molar mass of $PbO$:
$Pb: 207.2g $
$O: 16.00g$
207.2g + 16.00g = 223.2g
$ \frac{1 mole (PbO)}{ 223.2g (PbO)}$ and $ \frac{ 223.2g (PbO)}{1 mole (PbO)}$
2. The balanced reaction is:
$2PbS(s) + 3O_2(g) --\gt 2PbO(s) + 2SO_2(g)$
According to the coefficients, the ratio of $PbO$ to $PbS$ is 2 to 2:
$ \frac{ 2 moles(PbS)}{ 2 moles (PbO)}$ and $ \frac{ 2 moles (PbO)}{ 2 moles(PbS)}$
3. Calculate the molar mass for $PbS$:
Molar mass :
$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)}$
4. Use the conversion factors to find the mass of $PbS$
$128g(PbO) \times \frac{1 mole(PbO)}{ 223.2g( PbO)} \times \frac{ 2 moles(PbS)}{ 2 moles (PbO)} \times \frac{ 239.3 g (PbS)}{ 1 mole (PbS)} = 137g (PbS)$
