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How much sulfuric acid $(\ce{H2SO4})$ can be obtained from 5 kg of sulfur $\ce{S8}$?

answer: 15.312 kg

To make 1 molecules of $\ce{H2SO4}$ we need 1 atoms of S.

Now coming to $\ce{S8}$. This gives us 8 atoms of S. so from this we can obtain 8 $\ce{H2SO4}$ molecules if we take 1 atom of $\ce{S8}$ and use it to make $\ce{H2SO4}$ molecules. So,

  • 1 molecule of $\ce{S_8}$ $\ce{->}$ 8 molecules of $\ce{H2SO4}$
  • 0.019 moles of $\ce{S_8}$ $\ce{->}$ $8 \times 0.019=0.15$ moles of $\ce{H2SO4}$

Formula mass of $\ce{H2SO4} = 1(2)+32+16(4) = 98$
we know the formula
for $$\text{Number of moles} = \frac{\text{Mass}}{\text{Formula Mass}}$$

So from this $\text{Mass}_\ce{H2SO4} = \text{Moles} \times \text{Formula Mass} = 0.15 \times 98=14.7$

Now you see that my answer doesn't match to the answer. What did I do wrong?

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  • $\begingroup$ I'm also getting 14.7 kg$. Maybe the answer given is wrong $\endgroup$ – Binary Geek Apr 16 '15 at 5:26
  • $\begingroup$ Do you Think my method of doing this question is 100% correct? $\endgroup$ – anni Apr 16 '15 at 5:29
  • $\begingroup$ Yes it is. The stoichiometry and application of unitary method all look good to me. $\endgroup$ – Binary Geek Apr 16 '15 at 5:31
  • $\begingroup$ then our answer is correct. $\endgroup$ – anni Apr 16 '15 at 5:32
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    $\begingroup$ Too much approximation, both in your answer and the answer provided. $8\cdot0.019$ is not equal to $0.15$… If you pile up approximation, your result will start to diverge. In their answer $15.312$, they also used approximation with $M(\ce{O})=16$ instead of $15.999$ for instance, and $M(\ce{S})=32$ instead of $32.06$. The best answer given are indeed the one from ringo and Klaus. $\endgroup$ – Martigan Apr 16 '15 at 8:19
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$\ce{S6}$, $\ce{S8}$, $\ce{S12}$ – does it make a difference or is it just a trick to make the question more complicated than it is?

What you know for sure is:

  • $M(\ce{S}) = 32.065\ \mathrm{g}\cdot \mathrm{mol}^{-1}$
  • $M(\ce{H2SO4}) = 98.079\ \mathrm{g}\cdot \mathrm{mol}^{-1}$

$\frac{5000\,\mathrm{g}}{32.065\,\mathrm{g}\cdot \mathrm{mol}^{-1}} = 155.93\,\mathrm{mol}$

$155.93\,\mathrm{mol} \cdot 98.097\,\mathrm{g}\cdot \mathrm{mol}^{-1} = 15.29\,\mathrm{kg}$, which is close enough to the answer.

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I prefer simple dimensional analysis as a method of arriving at the answer. I'm sure it will simplify many of your future calculations.

${5 \hspace{.5mm} kg \hspace{1mm} S_8}·\frac{1000 \hspace{.5mm} g}{1 \hspace{.5mm} kg}·\frac{1\hspace{.5mm}mol\hspace{1mm}S_8}{256.528\hspace{.5mm}g\hspace{1mm}S_8}·\frac{8\hspace{.5mm}mol\hspace{1mm}H_2SO_4}{1\hspace{.5mm}mol\hspace{1mm}S_8}·\frac{98.07948\hspace{.5mm}g\hspace{1mm}H_2SO_4}{1\hspace{.5mm}mol\hspace{1mm}H_2SO_4}·\frac{1\hspace{.5mm}kg}{1000\hspace{.5mm}g}=15.2934\hspace{.5mm}kg\hspace{1mm}H_2SO_4$

The discrepancy in the answer is likely as a result of me using very precise values for the molar mass of the atoms used in the calculation, but clearly the answer is correct.

If you are not familiar with dimensional analysis, the goal is to cancel the units via dividing with fractions. You will notice that all the units in the above equation cancel to leave us with our desired answer, with the units $kg\hspace{1mm}H_2SO_4$.

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  • $\begingroup$ Expressions for units shall contain nothing else than unit symbols and mathematical symbols. Write ‘$\mathrm{kg}$’, not ‘$\mathrm{kg}\ \ce{H2SO4}$’. $\endgroup$ – Loong Apr 16 '15 at 8:42
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    $\begingroup$ @Loong yes, that is an SI guideline. And those guidelines were written with physicists and engineers in mind, not chemists, who often have to deal with several quantities of different substances, all with the same unit. Ask yourself: Would deleting the H2SO4 at the end make this answer more clear or less clear? Would you advocate deleting all the species from the working, too, which would lead to it being indecipherable? $\endgroup$ – Level River St Apr 16 '15 at 17:32

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