Compounds can only be cancelled if they are on

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MnO’s on the product side, we need 12 MnO’s on the reactant side. Compounds
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can only be cancelled if they are on opposite sides of the arrow. Therefore we must do the following to the third equation: 12(MnO(s) + ½ O 2 (g) MnO 2 (s) ) ∆H = 12(-136) Now, to get ∆H for the bottom reaction, add the ∆H values of the three equations: ∆H = (2)464kJ + 3(142kJ) + 12(-136kJ) = -278 kJ NOTICE: Once one element or compound is arranged correctly in an equation, all of the other will also be correct. There is no reason to try to arrange each compound and element in every equation. This final step is not necessary, but to prove the three equations add up to the final equation: 2(2Mn 3 O 4 (s) 6MnO(s) + O 2 (g)) ∆H = 2(464J) 3(4MnO 2 (s) 2 Mn 2 O 3 (s) + O 2 (g)) ∆H = 3(142J) 12(MnO(s) + ½ O 2 (g) MnO 2 (s) ) ∆H = 12(-136J) Multiplying through: 4 Mn 3 O 4 (s) 12MnO(s) + 2O 2 (g) 12MnO 2 (s) 6Mn 2 O 3 (s) + 3O 2 (g) 12MnO(s) + 6O 2 (g) 12MnO 2 (s) Canceling out species on opposite sides we get: 4Mn 3 O 4 (s) + O 2 (g) 6Mn 2 O 3 (s) ∆H = -278 kJ ( by adding each ∆H above)
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