Finding the amount of nitrous oxide produced in the reaction between lead sulfide and nitric acid

Question

Lead(II) sulfide dissolves in excess nitric acid according to the equation below. Calculate the volume of $\ce{NO{(g)}}$ at $\pu{27 ^\circ C}$ and $\pu{1.10 atm}$ produced from $\pu{4.7 g}$ of $\ce{PbS(s)}$.

$$\ce{3PbS{(s)} + 2NO3–{(aq)} + 8H+{(aq)} -> 3Pb^2+{(aq)} + 3S{(s)} + 2NO{(g)} + 4H2O{(l)}}$$

a) $\pu{0.29 L}$
b) $\pu{0.44 L}$
c) $\pu{0.66 L}$
d) $\pu{30 L}$
e) $\pu{45 L}$

I used the ideal gas formula and converted the mass of $\ce{PbS}$ into number of moles of $\ce{PbS}$ however got the answer b) $\pu{0.44 L}$ which is incorrect. The correct answer is a) however I am unsure about. Why this is?

Answer 1

The first step is to calculate the number of moles of $\ce{PbS}$:

$$n(\ce{PbS}) = \frac{\pu{4.7 g}}{\pu{239 g mol-1}} = \pu{0.0197 mol}$$

The next step is to calculate how many moles of $\ce{NO}$ will be produced based on the stoichiometry of the reaction. As stated in a comment, it looks like this is the step that was left out of your calculations. Because only $\pu{2 mol}$ of $\ce{NO}$ are produced for every $\pu{3 mol}$ of $\ce{PbS}$, only $2/3$ of $\pu{0.0197 mol}$, or $\pu{0.0131 mol}$ $\ce{NO}$ will be formed.

At this point we know all of the variables required to invoke the ideal gas law:

$$ \begin{align} V(\ce{NO}) &= \frac{nRT}{p}\\ &= \frac{\pu{0.0131 mol} \times \pu{0.0821 L atm mol-1 K-1} \times \pu{300K}}{\pu{1.10 atm}}\\ &= \pu{0.29 L} \end{align} $$

And this gives us the correct answer of a. Again, as your answer was off by a factor of $2/3$, it is pretty clear that the second step above regarding the stoichiometry of the reaction is the only step that you left out of your calculations.