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$\pu{48g}$ of graphite and $\pu{16g}$ of $\ce{O2}$ is given. The question is to count the no of molecules of $\ce{CO2}$ produced?

Now, the solution in book is this

No of $\ce{C}$ atoms = $\frac{\pu{48g}}{\pu{12g}}$ = $\pu{4mol}$. No of $\ce{O}$ atoms = $\frac{16}{16}$ =$\pu{1 mol}$ , and no of $\ce{O2}$ atoms = $\pu{0.5 mol}$.

The reaction given is

$\ce{C + 2O ->CO2}$.

$\ce{CO2}$ produced is $\pu{0.5 mol}$ here.

My question is why not to use the reaction $\ce{C + O2 ->CO2}$?

What would the difference be in the answer ?

S.M.T
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    If a chemistry textbook in this day and age uses terms and symbols such as "gm", "gms", "gmd" "no of X" and proposes reaction with atomic oxygen out of blue, it's not a chemistry textbook. It is something that may have an appearance or a title of a chemistry textbook, but it is not one. Probably a reprint from BuzzFeed or a comic book, but definitely not a chemistry textbook. – andselisk Feb 24 '21 at 17:41
  • @andselisk I am sorry for that. But it would be more helpful if you help me out here ? – S.M.T Feb 24 '21 at 17:47
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    @Sarabsrimt - The point is that we'd like chemistry problems to stay within the bounds of reality as much as possible to teach real chemistry as well as being an exercise. In this case you can't go to your chemical supplier and buy a tank of oxygen atoms. You must buy a tank of oxygen molecules. – MaxW Feb 24 '21 at 17:53
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    @Sarabsrimt It's not your fault, just avoid textbooks like this one. And yes, given the context your chemical reaction makes much more sense: $$\ce{C(s) + O2(g) -> CO2(g)}$$ The limiting reagent is oxygen: $$n(\ce{C}) = \frac{m(\ce{C})}{M(\ce{C})} = \frac{\pu{48 g}}{\pu{12 g mol-1}} = \pu{4 mol}$$ $$n(\ce{O2}) = \frac{m(\ce{O2})}{M(\ce{O2})} = \frac{\pu{16 g}}{\pu{32 g mol-1}} = \pu{0.5 mol}$$ and the answer is $$N(\ce{CO2}) = n(\ce{CO2})\times N_\mathrm{A} = n(\ce{O2})\times N_\mathrm{A} = \pu{0.5 mol}\times\pu{6.022E23 mol-1} \approx \pu{3.0E23}$$ – andselisk Feb 24 '21 at 18:08
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    @Sarabsrimt Actually, feel free to check community's assembly of Resources for learning Chemistry. The resources listed there are arguably one of the best. – andselisk Feb 24 '21 at 18:10
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    @andselisk Ok . Book I used was ICSE class 10 Viraj F Dalal. But I will surely check the resources. Thanks a lot. – S.M.T Feb 24 '21 at 18:18
  • @Sarabsrimt No prob, good luck and have a good one. – andselisk Feb 24 '21 at 18:19
  • @andselisk Are both reaction correct to use ? – S.M.T Feb 25 '21 at 04:35
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    @Sarabsrimt No, $\ce{C + 2 O -> CO2}$ doesn't make sense unless there is extra context provided. As MaxW already told you, oxygen doesn't casually dwell in atomic form. – andselisk Feb 25 '21 at 04:37
  • @andselisk Does C + 2O —> CO2 happen like this : C + O —> CO and then CO + O —> CO2 – S.M.T Feb 25 '21 at 04:37
  • @andselisk Ok. One more thing. Someone told me that these reaction only tell us how much are ingredients produced and not specifically tells if the reaction will happen or not. Is it true ? – S.M.T Feb 25 '21 at 04:38
  • @andselisk Right. So reaction isn’t possible but does it mean it tells us the amount of reactant or product that would be needed ? – S.M.T Feb 25 '21 at 04:39
  • @Sarabsrimt Stoichiometry. – andselisk Feb 25 '21 at 06:09

1 Answers1

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16 g of dioxygen contains 16 g of oxygen atoms. To find out how much carbon dioxide is produced, you can either use a realistic description of the reaction, or you can go atom by atom. Either path gives the same solution, but the one chosen by the textbook is not modern chemistry, and makes some of us angry or sad.

My question is why not to use the reaction $\ce{C + O2 -> CO2}$?

You should because that chemical equation describes the reactants and products much better than what the textbook has. It does not describe the mechanism (i.e. the steps to arrive at that net reaction), though.

What would the difference be in the answer.

Same numerical answer (unless you get confused because the reaction does not match the actual reactants).

Karsten
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