# Determining amounts of product and excess reagent remaining post reaction

$$\ce{2FeS2 + 11/2 O2 -> Fe2O3 + 4 SO2}$$

Mass of $$\ce{FeS2}$$ is $$\pu{600 g}$$ and mass of $$\ce{O2}$$ is $$\pu{800 g}$$. Find the amounts of $$\ce{Fe2O3}$$ and $$\ce{SO2}$$ and the remaining amount of the excess reagent.

### My Attempt

$$\frac{n_\ce{FeS2}}{2} = \frac{600/120}{2} = \frac{5}{2} \lt \frac{n_\ce{O2}}{11/2} = \frac{800/32}{11/2} = \frac{50}{11}\implies \ce{FeS2}~\text{is limiting reagent}$$

By Mole-Mole Analysis, we have

$$\frac{n_\ce{FeS2}}{2} = \frac{n_\ce{Fe2O3}}{1} = \frac{n_\ce{SO2}}{4}\implies \begin{bmatrix}n_\ce{Fe2O3} \\ n_\ce{SO2}\end{bmatrix} = \begin{bmatrix}5/2 ~\text{mol} \\ 10~\text{mol}\end{bmatrix}$$

$$\text{Excess reagent remaining} = \frac{800}{32} - \dfrac{5}{2}\cdot\dfrac{11}{2} = \pu{11.25 mol}$$

However, my answer does not match with my other classmates. What am I doing wrong?

• Were your classmates converting the answers to grams instead of moles? // IUPAC went to this crazy idea that you don't calculate the number of moles, but rather the "amount of a substance." I'm sure that a lot of "old" books that don't follow IUPAC's syntax used the word "amount" in a number of problems. – MaxW Apr 21 at 16:46
• @MaxW As an adept of the "crazy idea" you mentioned, I edited the "amount" term in as a more appropriate one. Guess what, originally there were good old "numbers of moles" (poor animals). – andselisk Apr 21 at 16:52
• @andselisk - If you're going to edit a problem like that I'd suggest that you note the change and the reason for it at the bottom of the problem. A lot of the questions seem to be posed by students from India who are using books with old condiations for STP (ie 22.4 liters/mole @STP) and "number of moles" instead of "amount." – MaxW Apr 21 at 17:11
• @andselisk, "But yeah, that's a good point, I have very little clue about chemical education in India" ...There is huge variation from place to place from very obsolete to quite modern. It is catching up I guess. Depends on the university. Some IITs are as good as Ivy League schools. Recently, I heard an Indian PhD student say "nascent hydrogen" does the reduction. This was a concept taught in 1940s. Similarly normality is still taught, as well as Sommerfeld model, parachor values (if you ever heard of it!) etc. Hard to generalize. – M. Farooq Apr 21 at 17:38
• Thanks for taking part in the discussion and resolving my issue. @andselisk Cheers :) – Paras Khosla Apr 22 at 7:43

Note the following relations from the equation you wrote: 1 mol $$\ce{FeS2}$$ = (11/4) mol $$\ce{O2}$$; 1 mol $$\ce{FeS2}$$ = (1/2) mol $$\ce{Fe2O3}$$; 1 mol $$\ce{FeS2}$$ = 2 mol $$\ce{SO2}$$.

You already determined that $$\ce{FeS2}$$ is the limiting reactant because we have 600/120 mol of iron sulfide and 800/32 mol of $$\ce{O2}$$;

This implies that 5 mol $$\ce{FeS2}$$ will require =5x (11/4) mol $$\ce{O2}$$.

Remaining mol oxygen = 25-13.75=11.25

Now use the relations established above: 5 mol iron sulfide would produce (5/2) or 2.5 mol iron oxide, and 5 mol iron sulfide would produce 5x2 = 10 mol sulfur dioxide.

• But this is pretty much what OP did, isn't it? – andselisk Apr 21 at 16:23
• True but just for confirmation only-parsed it for his classmates who are doing it wrong :-) The boy did a very good job in solving this relatively difficult problem for a 10th grader. – M. Farooq Apr 21 at 16:43
• OP did very well indeed, how about showing him support with upvotes?;) – andselisk Apr 21 at 16:56
• Slight off topic Q: How does this reputation matter in StackExchange? I noted some people across the board (not just chem) have very high scores. Is there any commercial aspect behind that? The scoring system became an issue in ResearchGate (I am sure you know it). They finally removed upvote or downvote option. After that some academics made an issue that some universities in developing countries use RG score for faculty promotions. In short how how does this reputation score help or is it just for self-satisfaction? In RG nobody can be anonymous. They need official academic address. – M. Farooq Apr 21 at 17:13
• UPVOTING - @andselisk I consider that up voting the question indicates that it is an interesting question that I'd like to see answered too, rather than giving the OP a pat on the back. In other words for these homework problems there isn't a way to vote for the poster's answer. – MaxW Apr 21 at 17:22

Given the problem as stated, you did the problem correctly. However I would have done it a bit differently to make it easier to check. I like to do the problems in steps. I also dislike carrying a lot of fractions in intermediate calculations since I get confused easily. (This is basically M. Farooq's answer with some explanation.)

### Given

$$\ce{2FeS2 + 11/2 O2 -> Fe2O3 + 4 SO2}$$

Mass of $$\ce{FeS2}$$ is $$\pu{600 g}$$ and mass of $$\ce{O2}$$ is $$\pu{800 g}$$. Find the amounts of $$\ce{Fe2O3}$$ and $$\ce{SO2}$$ and the remaining amount of the excess reagent.

### Molar Quantities

For $$\ce{Fe}$$:

$$n_\ce{FeS2} = \frac{600}{120.} = 5.0000$$

For $$\ce{O2}$$

$$n_\ce{O2} = \frac{800}{32.0} = 25.000$$

Here I calculate the decimal amounts rather than carrying fractions. Since it would seem that the problem has 3(?) significant figures, I carried two extra for the intermediate calculation.

### Stoichiometric Quantities

For $$\ce{Fe}$$:

$$\frac{n_\ce{FeS2}}{2} = \frac{5.0000}{2} = 2.5000$$

For $$\ce{O2}$$

$$\frac{n_\ce{O2}}{11/2} = \frac{25.000}{11/2} = \frac{50.000}{11} =4.4545$$

Thus $$\ce{FeS2}$$ is limiting reagent

Here I could have reached the conclusion in my head, but I do the math anyway so that I don't make a stupid mistake and also so that I can check the problem easily.

### Rewrite the chemical equation

Since we now know that $$\ce{FeS2}$$ is the limiting reagent, rewrite the chemical equation in terms of $$\ce{FeS2}$$.

$$\ce{FeS2 + 11/4 O2 -> 1/2Fe2O3 + 2 SO2}$$

Now for each mole of $$\ce{FeS2}$$:

• You use 11/4 mole $$\ce{O2}$$ so $$\ce{O2}$$ remaining is:

$$25.0000 - (5.0000*11/4) = 25.0000 - 13.75 = 11.25$$

• You get 1/2 mole $$\ce{Fe2O3}$$ so:

$$5.0000 / 2 = 2.50$$

• You get 2 mole $$\ce{SO2}$$ so:

$$5.0000 * 2 = 10.0$$

Now for a final check I note that at the start there were 25.0 moles of $$\ce{O2}$$, but I have only 11.25 left. That is good since some was consumed. If I had ended up with more oxygen I obviously would have a mistake...

• LOL - When I started doing these problems in high school I had a slide rule which was a step up from using 3 place log tables. I didn't a calculator until I was in grad school. – MaxW Apr 21 at 18:40
• I have not used slide rule but my boss showed slide rule with pride. However I have used log tables in late 1990s :-). The trouble of finding characteristic and mantissa from printed tables with fine print ...good old memories. How many students even know what a mantissa is. – M. Farooq Apr 21 at 18:47
• @M.Farooq - Gave my Post Versalog (which was a professional tool, not a toy) slide rule to my sister who wanted it to play with. I have thought since that it would be nice to have it in a shadow box on the wall. The glass on the shadow box would read: In case of power failure, break glass. – MaxW Apr 21 at 18:52
• Good one. As I told someone yesterday who was asking here how I survived without learning LaTex for writing papers. My answer was that very good science was also done when we had typewriters. Slide rule produced prodigies and geniuses, this was the time when Americans and Russians were also going to the moon. They had computers though, I believe. – M. Farooq Apr 21 at 19:00