25.9 kJ + 1/2 H2 + 1/2 I2 -> HI

Question

B) how much energy is needed for the reaction of 4.57E24 molecules of iodine, I2, with excess hydrogen, H2?

How does the "excess" hydrogen come into play? I mean, what's the significance of having excess H2 and how do I figure out the answer considering that H2 is in excess? What does it change?

One more quick question!:
If, for example an equation is P4O10 + 6H2O -> 4H3PO4 + 257.2 kJ and the question is, "how much energy is released when 235 g of H3PO4 is formed?" then does that mean there's 235 g in ONE Mole of H3PO4?

Answer 1

When an "excess" is used, it simply means that the reaction will use what is needed. It means that the "other" reactant is the limiting reagent.
4.57E24 molecules I2 = 4.57E24/6.02E23 = mols I2 = about 7.6 mols or about 1900 grams (not exact). 1/2 that is about 900g I.
So 25.9 kJ x (900/126.9) = heat absorbed.

b.
No, you're making up numbers. Read the equation for what it is. That equation means that 257.2 kJ energy is released WHEN THAT REACTION OCCURS; i.e., when 1 mol P4O10(283.9g) reacts with 6 mols H2O(108g) to produce 4 mol H3PO4(392g).