Calculating Net Moles

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Rhiannon Imbeah 2I
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Calculating Net Moles

Postby Rhiannon Imbeah 2I » Tue Jul 05, 2016 5:07 pm

In one of the post-assessment question regarding balancing chemical equation, we are asked to find the net number of moles produced. How would you go about finding the net number of moles produced in a chemical reaction? Thank you

Chem_Mod
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Re: Calculating Net Moles

Postby Chem_Mod » Tue Jul 05, 2016 5:19 pm

If the amounts for reactants are provided, find the limiting reactant first. Then use the stoichiometric ratio shown in the balanced equation between the limiting reactant to the product to find the net moles of product produced.

Drake_Everlove_1K
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Re: Calculating Net Moles

Postby Drake_Everlove_1K » Wed Sep 28, 2016 7:35 pm

I wanted to follow up on this question; I too am hung up on one of the pre/post-assessment questions. It does not seem to be explained in the video module. Here is the question:

8. During a summer camping weekend 4 moles of butane (C4H10) gas were used for cooking. Chose the right balanced equation for the combustion of 4 moles of butane gas. What is the net number of moles of gas produced?

A. 4C4H10(g) + 26O2(g) → 16CO2(g) + 20H2O(g); 6 (my answer)

B. 4C4H10(g) + 26O2(g) → 16CO2(g) + 18H2O(g); 4

C. 4C4H10(g) + 25O2(g) → 16CO2(g) + 20H2O(g); 6

D. 4C4H10(g) + 26O2(g) → 16CO2(g) + 20H2O(g); 5

Now, based on intuition, I imagine that the number of moles of gas produced by this reaction is six, because if you add the total number of moles of atoms (represented by the stochiometric coefficients) on each side, you end up with 30 moles on the left and 36 moles on the right, a difference of six.

It almost seems like that goes against the law of conservation, but I can only assume the total grams that each mole of atoms represent are equal. Whatever mechanism may be involved, it does seem like six moles of atoms were "produced" to be on the right side. The way I got to this conclusion has nothing to do with finding the limiting reactant, as the above poster recommended, so I must be missing something and I hope this can be explained to me.

DavidEcheverri3J
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Re: Calculating Net Moles

Postby DavidEcheverri3J » Tue Oct 02, 2018 12:20 pm

Drake_Everlove_1K wrote:I wanted to follow up on this question; I too am hung up on one of the pre/post-assessment questions. It does not seem to be explained in the video module. Here is the question:

8. During a summer camping weekend 4 moles of butane (C4H10) gas were used for cooking. Chose the right balanced equation for the combustion of 4 moles of butane gas. What is the net number of moles of gas produced?

A. 4C4H10(g) + 26O2(g) → 16CO2(g) + 20H2O(g); 6 (my answer)

B. 4C4H10(g) + 26O2(g) → 16CO2(g) + 18H2O(g); 4

C. 4C4H10(g) + 25O2(g) → 16CO2(g) + 20H2O(g); 6

D. 4C4H10(g) + 26O2(g) → 16CO2(g) + 20H2O(g); 5

Now, based on intuition, I imagine that the number of moles of gas produced by this reaction is six, because if you add the total number of moles of atoms (represented by the stochiometric coefficients) on each side, you end up with 30 moles on the left and 36 moles on the right, a difference of six.

It almost seems like that goes against the law of conservation, but I can only assume the total grams that each mole of atoms represent are equal. Whatever mechanism may be involved, it does seem like six moles of atoms were "produced" to be on the right side. The way I got to this conclusion has nothing to do with finding the limiting reactant, as the above poster recommended, so I must be missing something and I hope this can be explained to me.

I think you might be right. My reasoning is that there are different numbers of atoms in a molecule of C4H10 and O2 and molecules of H2O and CO2, therefore, the law of conservation of mass would still allow for the 6 moles to be produced. The only thing occurring is the molecules are changing the elements and atoms that make them up, therefore creating different net moles, but there would still be the same number of atoms on either side. I had no idea how this worked until I saw your post and I just had a Eureka! moment. Thanks!


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