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

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?

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).

B) In this reaction, the hydrogen, H2, is considered to be in excess, which means there is more than enough of it to react with the iodine, I2. The significance of having excess H2 is that it ensures that all the iodine will react and none will be left over. So, in this case, the amount of energy needed for the reaction will be determined by the amount of iodine, I2, because it is the limiting reactant (the one that determines the maximum amount of product that can be formed). To figure out the answer, you will need to use the balanced equation to determine the stoichiometric ratio between iodine, I2, and energy (kJ) in the reaction.

For the equation P4O10 + 6H2O -> 4H3PO4 + 257.2 kJ, the energy released refers to the energy released per mole of H3PO4 formed. So, if you are asked how much energy is released when 235 g of H3PO4 is formed, you will need to convert the mass of H3PO4 to moles using the molar mass of H3PO4 and then use the stoichiometric ratio in the balanced equation to determine the amount of energy released. It does not necessarily mean that there is 235 g in one mole of H3PO4, but rather the question is asking for the energy released when a specific mass of H3PO4 is formed.

To find out how much energy is needed for the reaction of 4.57E24 molecules of iodine, I2, with excess hydrogen, H2, we need to calculate the energy needed for the given number of molecules.

First, let's start by finding the moles of iodine, I2:
Molar mass of I2 = 2 * atomic mass of iodine = 2 * 126.9 g/mol = 253.8 g/mol

Moles of I2 = (mass of I2) / (molar mass of I2)
= (4.57E24 molecules) / (Avogadro's number)
= (4.57E24) / (6.022E23 mol^-1)
≈ 7.59 mol

Next, let's look at the equation: 25.9 kJ + 1/2 H2 + 1/2 I2 -> HI

The coefficients in front of H2 and I2 indicate the stoichiometric ratio. In this case, 1/2 H2 and 1/2 I2 mean that the reaction requires equal amounts of hydrogen and iodine to form HI.

However, the problem states that there is an excess of hydrogen. This means that there is more than enough hydrogen available for the reaction, and it will not be fully consumed. As a result, the amount of hydrogen does not play a role in determining how much energy is needed.

To calculate the energy needed for the reaction, we only need to consider the moles of iodine (7.59 mol in this case). The energy needed can be calculated by multiplying the moles of iodine by the energy given in the equation.

Energy needed = 7.59 mol * 25.9 kJ/mol
= 198 kJ (rounded to the nearest whole number)

Therefore, the energy needed for the reaction of 4.57E24 molecules of iodine with excess hydrogen is 198 kJ.

Regarding your quick question about the equation P4O10 + 6H2O -> 4H3PO4 + 257.2 kJ:

In the given equation, the reaction produces 4 moles of H3PO4 for every 1 mole of P4O10. However, the amount of H3PO4 mentioned in the question (235 g) is not necessarily one mole.

To determine the number of moles from the given mass, you need to divide the mass by the molar mass of H3PO4.

Molar mass of H3PO4 = (3 * atomic mass of H) + atomic mass of P + (4 * atomic mass of O)
= (3 * 1.01 g/mol) + 31.0 g/mol + (4 * 16.0 g/mol)
= 98.0 g/mol

Moles of H3PO4 = (mass of H3PO4) / (molar mass of H3PO4)
= 235 g / 98.0 g/mol
≈ 2.40 mol

From this, you can see that 235 g of H3PO4 corresponds to approximately 2.40 moles.

B) To determine how much energy is needed for the reaction of 4.57E24 molecules of iodine with excess hydrogen, you will need to calculate the energy change per mole of iodine and then multiply it by the number of moles of iodine present in 4.57E24 molecules.

1. First, find the energy change per mole of iodine. The given reaction has a ΔH value of 25.9 kJ, which means it releases 25.9 kJ of energy when 1 mole of iodine reacts.

2. Next, convert the number of molecules of iodine (4.57E24) to moles. Since Avogadro's number is approximately 6.022 x 10^23, you can divide the number of molecules by Avogadro's number to get the number of moles.

Moles of iodine = 4.57E24 / (6.022 x 10^23)

3. Finally, multiply the number of moles of iodine by the energy change per mole to find the total energy needed:

Energy needed = Moles of iodine * ΔH

The "excess" hydrogen in this case means that there is more than enough hydrogen available for the reaction. It is not a limiting reactant, and therefore, its quantity does not affect the amount of energy needed for the reaction. You only need to consider the amount of iodine given and calculate the energy change based on that.

For the second question:

If the equation is P4O10 + 6H2O -> 4H3PO4 + 257.2 kJ, and you want to find out how much energy is released when 235 g of H3PO4 is formed, you need to follow these steps:

1. Determine the number of moles of H3PO4 using its molar mass. The molar mass of H3PO4 is approximately 98 g/mol.

Moles of H3PO4 = Mass of H3PO4 / Molar mass of H3PO4

2. Finally, multiply the moles of H3PO4 by the energy change per mole (given as 257.2 kJ) to find the total energy released:

Energy released = Moles of H3PO4 * ΔH

To clarify your last question, the 235 g of H3PO4 is not specifically related to one mole of H3PO4. The molar mass of H3PO4 (98 g/mol) tells you that 1 mole of H3PO4 weighs approximately 98 grams. If you have 235 grams of H3PO4, you can determine the number of moles using the molar mass, as explained in step 1 above.