I'm having minor difficulties solving this problem:

How many grams of sulfur (VI) oxide can be derived from 8 g of sulfur and 6 g of oxygen? (S-32)

I've tried to solve it this way:

$\ce{2S + 3O2 ->2SO3}$

$\ce{S}$ = 8g : 32 g/mol = 0.25 mol

$\ce{O2}$ = 2 x (6 g : 32 g/mol) = 0.375 mol

Neither oxygen nor sulfur is in surplus, therefore:

$\ce{S}$ = $\ce{SO3}$

$\ce{SO3}$ = 80 g/mol

$\ce{SO3}$ = 80 g/mol x 0.25 mol = 20 g

But the correct answer is 10 g of $\ce{SO3}$. What did I do wrong?

  • $\begingroup$ You write an equation with O2, and then use n(O). That's a contradiction. $\endgroup$ – Ivan Neretin May 20 '18 at 16:09
  • $\begingroup$ I've corrected it. I thought that only one atom of oxygen is relevant. $\endgroup$ – Aleksa May 20 '18 at 16:13
  • $\begingroup$ What would you do with the other atom? Of course both are equally relevant. Now your n(O2) is wrong. $\endgroup$ – Ivan Neretin May 20 '18 at 16:22
  • $\begingroup$ Is it correct now? $\endgroup$ – Aleksa May 20 '18 at 16:27
  • $\begingroup$ No. Why the "$2\times$"? $\endgroup$ – Ivan Neretin May 20 '18 at 17:38

The way to approach these limiting reagent stoichiometry reactions, as they are referred to, is to first determine the molar quantities of each of the reagents. You have correctly done so for sulfur, but not for diatomic oxygen. There is no reason, as Ivan Neterin has pointed out, to multiply by 2 the number of moles of oxygen. Doing this will in effect, cancel out the 32 g/mol of $\ce{O2}$ and you will solve this problem like there is 6 g monatomic oxygen [$\ce{O}$], rather than what there really is [diatomic oxygen [$\ce{O2}$]].

The next step is to set up a sort of "need v. have" proportion in order to determine which reagent is the limiting reagent.

Since there are .25 moles of $\ce{S}$ and .1875 moles of $\ce{O2}$, we can determine the limiting reagent is $\ce{O2}$ This is because, if sulfur was the limiting reagent, we would need greater than 1.5 times the current quantity of $\ce{O2}$
In essence, if $\ce{S}$ was the limiting reagent, the reaction would be $\ce{.25S + .375 O2->.25SO3}$,
while if $\ce{O2}$ was the limiting reagent, the reaction would be $\ce{.125S + .1875 O2 ->.125 SO3}$

As you can see, when $\ce{S}$ is the limiting reagent, we write too much $\ce{O2}$ than we really have [our reaction says we have .375 moles $\ce{O2}$ when we really have .1875 moles$\ce{O2}$]. In the second reaction, we have more than enough $\ce{S}$ than the reaction calls for. Therefore, $\ce{.125 moles or ~10 g SO3}$ is produced.

I believe your problem was converting the $\ce{O2}$ to moles; you must remember that the molar mass of the compound is 32, and you have 6 g of that compound, not double 6, or 12 g.

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