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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?

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  • $\begingroup$ 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
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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.

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