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Enthalpy change question??? This is what I have so far but I'm confused about the mole...

Enthalpy change question??? This is what I have so far but I'm confused about the mole part of the question. P4(s) + 10 Cl2(g) = 4PCl5(s) ΔH(rxn) = -1774.0 kJ 4 PCl5(s) = 4 PCl3(l) + 4 Cl2(g) ΔH(rxn) = 495.2 kJ --------------------------------------… ----------------------------------------… -------- P4(s) + 6 Cl2(g) = 4 PCl3(l) ΔH(rxn) = [-1774.0 kJ + 495.2 kJ] = -1278.8 kJ "Use these data to calculate the enthalpy change for the formation of 3.50 mol of from phosphorus and chlorine." How do I do this?
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Answer #1

P4(s) + 10Cl2(g) 4PCl5(s)           ΔH = -1774.0 kJ

4PCl5(s) 4PCl3(l) + 4Cl2(g)         ΔH = 495.2 kJ

P4(s) + 6Cl2(g) = 4PCl3(l)                ΔH = [-1774.0 kJ + 495.2 kJ] = -1278.8 kJ

Enthalpy change for the formation of 4 moles of PCl3 from phosphorus and chlorine = -1278.8 kJ

So, Enthalpy change for the formation of 3.50 moles of PCl3 from phosphorus and chlorine = (-1278.8/4) 3.50 = -1118.95 kJ

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