The Question:

How many liters of $H_2S$ would be required to react with excess oxygen to produce $12.0$ $L$ of $SO_2$? The reaction yield is $88.5$%. Assume constant temperature and pressure throughout the reaction.

My Approach:

I used dimensional analysis to convert from liters of $SO_2$ to $moles$ of $H_2S$ and then used Avogadro`s Law ($\frac{V_1}{n_1} = \frac{V_2}{n_2}$) to find the volume of $H_2S$ in liters.

Dimensional Analysis: $\frac{1}{.885} * \frac{12.0L}{1} * \frac{1 g}{1 mL} * \frac{1 mol SO_2}{64g} * \frac{2 mol SO_2}{2 mol H_2S} * \frac{1 mol H_2S}{34 g} = 6.23$ $mol$ $of$ $H_2S$.

Avogadro`s Law: $\frac{12.0L}{166 mol} * \frac{ L }{6.23 mol}$

End Result = $.028 L$

^ The end result does not seem right and any help in remedying my approach or shedding light on another approach would be much appreciated. Thanks in advance!

  • $\begingroup$ Is the reaction $\ce{H_2S}+ \frac{3}{2}O_2 ->H_2O+SO_2$? $\endgroup$ – Sherlock Holmes Oct 19 '14 at 1:33
  • $\begingroup$ Is it given that the density of $\ce{SO_2}$ is 1 gram per mL? As it's a gas I think it would be much lower. $\endgroup$ – Sherlock Holmes Oct 19 '14 at 1:45

I never really learned the method of solving this stuff with Avogadro's Law, but here's an almost identical way.

Your balanced equation is as follows...

$\ce{2H2S + 3O2 -> 2SO2 + 2H2O}$, and I think you agree on that by looking at your calculations.

At a given temperature and pressure, $1 mol = x L$. For example, at STP (standard temperature and pressure) this is $1 mol = 22.4 L$. It wasn't exactly defined what temperature and pressure are, but they are constant, so we can use a variable $x$ to denote some conversion (which cancels out anyway, I assume giving rise to Avogadro's Law).

$12.0L_{\ce{SO2}} * \frac{1mol_{\ce{SO2}}}{xL_{\ce{SO2}}} * \frac{2mol_{\ce{H2S}}}{2mol_{\ce{SO2}}} * \frac{xL_{\ce{H2S}}}{1mol_{\ce{H2S}}} = 12.0L_{\ce{H2S}}$

But this is assuming $100\%$ yield. I see you divided by $0.885$ above, which is correct, but I've gotten a metric ton of these wrong before, so I prefer the proportions way.

Regardless, $\frac{12L_{\ce{H2S}}}{0.885} = 13.6L_{\ce{H2S}}$

I think you definitely overcomplicated things by assuming a density of $1g/mL$ among other things. Your dimensional analysis also seems strange near the end of the left hand side where you somehow cancel $g$ with $mol_{\ce{SO2}}$.

If I were to do it with Avogadro's Law (taking a guess here), I'd still get $12.0L$ because the mole ratio is the same. And then I'd divide like you did.


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