Asked by Anonymous
The amount of energy released by the combustiom of 100 g of C2H2 (g) is ______ MJ.
2CH2 + 5O2 ----> 4CO2(g) + 2H2O(g) ΔH= -2511.0 kJ
Do I use ΔH = Hproducts - Hreactants? And if so how would that help me solve for the energy released?
2CH2 + 5O2 ----> 4CO2(g) + 2H2O(g) ΔH= -2511.0 kJ
Do I use ΔH = Hproducts - Hreactants? And if so how would that help me solve for the energy released?
Answers
Answered by
DrRebel
I'm assuming the CH2 should be C2H2 (Acetylene) ...
From the balanced equation, two moles of acetylene => -2511Kj = 2.511Mj
Convert 100g C2H2 to moles 100g/26g/mol = 3.85 moles
Set ratio ...
(2/3.85 = 2.511/x => x = [(2)(2.511)]/(3.85)Mj = 1.30Mj
From the balanced equation, two moles of acetylene => -2511Kj = 2.511Mj
Convert 100g C2H2 to moles 100g/26g/mol = 3.85 moles
Set ratio ...
(2/3.85 = 2.511/x => x = [(2)(2.511)]/(3.85)Mj = 1.30Mj
Answered by
DrBob222
No and yes.
You don't need to use dHrxn = (n*dHf products) - (n*dHf reactants) because the problems has already done that for you. For the reaction given dH is -2511.0 kJ and that is for 2 mols of C2H2 or 2*26 = 52 g.
So the revised question is if 52 g releases 2511.0 kJ how much will 100 g release? Can you take it from there?
You don't need to use dHrxn = (n*dHf products) - (n*dHf reactants) because the problems has already done that for you. For the reaction given dH is -2511.0 kJ and that is for 2 mols of C2H2 or 2*26 = 52 g.
So the revised question is if 52 g releases 2511.0 kJ how much will 100 g release? Can you take it from there?
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