Lead(II) sulfide dissolves in excess nitric acid according to the equation below. Calculate the volume of $\ce{NO{(g)}}$ at $27 ^\circ \mathrm{C}$ and $1.10\ \mathrm{atm}$ produced from $4.7\ \mathrm{g}$ of $\ce{PbS(s)}$. $$\ce{3PbS{(s)} + 2NO3^{–}{(aq)} + 8H+{(aq)} -> 3Pb2+{(aq)} + 3S{(s)} + 2NO{(g)} + 4H2O{(l)} }$$

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

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) $0.44\ \mathrm{L}$ which is incorrect. The correct answer is a) however I am unsure about. Why this is?

  • $\begingroup$ You skipped a step. You got the number of moles of $\ce{PbS}$, that's good. Now you need to find the number of moles of $\ce{NO}$ which result from the reaction, and only then apply the ideal gas law. $\endgroup$ – Nicolau Saker Neto May 22 '16 at 5:43

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

$$\mathrm{\frac{4.7g\ \ce{PbS}} {239g\ \ce{PbS}\ \ mol^{-1}\ \ce{PbS}} = 0.0197\ mol\ \ce{PbS}}$$

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 2 mols of $\ce{NO}$ are produced for every 3 mols of $\ce{PbS}$, only $\mathrm{\frac{2}{3}}$ of 0.0197 mol, or 0.0131 mol $\ce{NO}$ will be formed.

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

$$\mathrm{V = \frac{nRT}{P}}$$

$$\mathrm{V = \frac{0.0131\ mol\ \ce{NO} * 0.0821\ L\ atm\ mol^{-1} K^{-1} * (300K)}{1.10\ atm\ \ce{NO}}}$$

$$\mathrm{V = 0.29\ L\ \ce{NO}}$$

And this gives us the correct answer of a. Again, as your answer was off by a factor of $\mathrm{\frac{2}{3}}$, 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.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.