Okay, so for home work i was given this problem.

Suppose that 8.00g of CH4 reacts with 16.00g O2 to form CO2 and H2O, find the grams of CH4, H2o, and O2 left over

Balanced Equation : CH4 + 2O2 = CO2 + 2H2O

My Calculations for limiting Reagent:

8.00g CH4(1 mol CH4/16.04g CH4)(2 mol H2O/1 mol CH4)(18.01g/1 mol H2O) = 18.0g H2O

16.00g O2(1 mol O2/32.00g O2)(2 mol H2O/2 mol O2)(18.01g H2O/1 mol H2O) = 18.0g H20

So, if they both equal the same amount of H2O created, which one would be considered my liming reagent?

  • 3
    $\begingroup$ From your balanced reaction, you see that one mole of methane reacts with two moles of oxygen, so 16 g methane with 64 g oxygen. Therefore, 8 g methane requires 32 g oxygen, but you only have 16 g oxygen. $\endgroup$ – Ed V Mar 25 '20 at 17:45
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    $\begingroup$ 8g of methane is 0,5 mol. 16g of O2 is 0.5 mol. There are the same number of moles of both, however you need two times as many moles of O2 as you do of CH4 for the reaction (2:1 mole ratio). Which is the limiting reagent? $\endgroup$ – arevmelikyan Mar 25 '20 at 17:48
  • $\begingroup$ Please rewrite this expression : 16.00g O2(1 mol O2/32.00g O2)(2 mol H2O/*2* mol O2)(18.01g H2O/1 mol H2O) = 18.0g H20. And change the * 2 * by * 1 * $\endgroup$ – Maurice Mar 25 '20 at 17:49

Limiting reagent is the reactant used which produces lesser amount of the product as compared to other reactants if other reactants are used in sufficient amounts. Thus it limits the quantity of the product.

Now, according to your question,

Moles of CH4

= (Mass of CH4 taken)÷(Molecular mass of CH4)

= 8/[12+(4×1)] = 8/16

= 0.5 moles of CH4 are present


Moles of O2 = (Mass of O2 taken)÷(Molecular mass of O2)

= 16/(16×2) = 16/32

= 0.5 moles of O2 are present

The reaction given is

CH4 + 2O2 => CO2 + 2H2O


For 1 mole of CH4

=> Produces 1 mole of CO2 and 2 moles of H2O

This means that

For 0.5 mole of CH4

=> Produces 0.5 mole of CO2 and 1 mole of H20

For 2 moles of O2

=> Produces 1 mole of CO2 and 2 moles of H2O

This means that

For 1 mole of O2

=> Produces 0.5 mole of CO2 and 1 mole of H2O

This means that

For 0.5 mole of O2

=> Produces 0.25 mole of CO2 and 0.5 moles of H2O

Since, O2 produces the lesser amount of products, thus, it is the Limiting Reagent.

Therefore, CH4 will be left as it is the reagent in excess.

Thus, with every 2 moles of O2

1 mole of CH4 is used.

This means that

With every 1 mole of O2 0.5 mole of CH4 are used

With every 0.5 moles of O2 0.25 moles of CH4 are used.

This means that in the reaction, quantity of

  1. CH4 = (0.5–0.25) moles = 0.25 moles

  2. O2 = 0 mole

  3. CO2 = 0.25 moles

  4. H2O = 0.5 moles

Therefore mass of a compound

= (Moles of compound that are present) ÷ (Molecular mass of that compound)

**Mass of

  1. CH4 = 0.25 × 16 = 4 grams

  2. O2 = 0 × 32 = 0 gram

  3. CO2 = 0.25 × 44 = 11 grams

  4. H2O = 0.5 × 18 = 9 grams **

Sorry if there's any mistake. Thank you.

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    $\begingroup$ Your final answers were correct. But you have gone awfully long way to prove it. $\endgroup$ – Mathew Mahindaratne Mar 26 '20 at 4:51
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    $\begingroup$ Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Also, check the correctness of the unit symbols and avoid using "textual" formulas. Note that it is illiterate to replace a physical quantity with its units: the amount of substance is the proper term, not moles of X. And I agree with @MathewMahindaratne: the answer is unnecessarily over-complicated. $\endgroup$ – andselisk Mar 26 '20 at 6:15
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    $\begingroup$ @MathewMahindaratne, Well, I am a student, although I can do it in no more than a minute but I just wanted the answer to be clear to the one who has asked question, maybe that is why you feel that it was long. You may know the topic elaborately but it may not be the same for the person asking the question. That is why I wrote it in this way. But as I have read about you, you are a chemist and i am a great fan of people who do research (because I too want to do it in future), so I am very sorry that it wasn't according to you, but I'll take your words as advice that I'll follow. Thank you. $\endgroup$ – Sanya Mar 26 '20 at 15:23
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    $\begingroup$ @Sanya: Don't get too personal. I already up-voted your answer. My comment is for future perspectives. In future, avoid repeating same argument. :-) $\endgroup$ – Mathew Mahindaratne Mar 26 '20 at 15:29
  • $\begingroup$ @MathewMahindaratne:It's okay, I didn't took it in a personal way. $\endgroup$ – Sanya Mar 26 '20 at 15:37

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