Solving Chemistry Enthalpy Homework: Step-by-Step Guide

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The discussion focuses on calculating the enthalpy change for the reaction involving ethane (C2H6) converting to ethylene (C2H4) and hydrogen (H2). The enthalpy change (ΔH) for the reaction is given as -64.2 kJ for two moles of C2H6, leading to confusion about how to apply this to 6 grams of the substance. Participants clarify that to find the enthalpy for 6 grams, one must first determine the number of moles in that mass, using the molar mass of C2H6, which is approximately 30 grams per mole. The correct calculation yields an enthalpy change of -6.41 kJ when using the appropriate conversion factors. Understanding the relationship between grams, moles, and enthalpy is crucial for solving this type of chemistry problem.
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Homework Statement



Given 2C2H6 \rightarrow2C2H4(g) 2H2(G) \Delta H=-64.2kJ

\Delta H for the reaction of 6.000g of C2H6(g) to givev C2H4(g) andH2(g) is

Homework Equations


\Delta H = \Sigma products - \Sigma reactants
q=ct
keeping the units
The enthalpies are not given and are not to be looked up

The Attempt at a Solution


I understand that i have to divide ΔH by 2 and multiply by 6g
but that gives me a unit of kJ*g so i need to divide by grams to get the units right... my solution book tells me to divide by 30g but i am unaware as to where that comes from
 
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You are given the enthalpy for a reaction of one mole. How much does a mole weigh in grams of this material? That would be my thinking, but the book has a slightly different number than I would expect
 
Book is right.

BOYLANATOR said:
You are given the enthalpy for a reaction of one mole.

Enthalpy is given for two moles, not for one. But generally speaking you are on the right track, it is abut finding number of moles.
 
The answer from the book is-6.41 kJ. I found the mols of C2H6 and used the ratio to figure out the sum of mass of reaction and products to get to get 12.00g this gives me the answer of -16.05 kJ.
=-64.2/2 to get it to a 1 to 1 ratio
=-32.1*6 multiply by 6.00g
=-192.6/12
=-16.05 kJ
if you divide by 30 instead of 12 you get the right answer i just don't get where it comes from
 
I don't follow your logic here. You worked out the enthalpy would be -32.1 kJ for a one mole reaction. That's ok.

But in this case it's not one mole of C2H6, it's 6 grams. What fraction of a mole is 6 grams?
 
Just make sure that the units cancel properly:

(32.1 \frac{kJ}{mole})(\frac{1mole}{30grams})(6grams)=? kJ
 
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