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
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?
2 answers
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?