The standard enthalpy of formation of Al3-(aq) is -538.4kJ/mol. How much heat is evolved when 0.3724 g of Al(s) reacts with excess hydrochloric acid?

0.3724 g x 26.98MW = 10.05 mol

-538.4kJ x 10.05 mol = -5410 kJ/mol
5410 kJ/mol of heat is evolved?????

Do you mean Al^3+ and not Al^3-?

moles = grams/molar mass not grams x molar mass.
You need to write and balance an equation, also.

yes, Al3+ sorry. so the moles=.01380. Could you help with the equation?

To find out how much heat is evolved when 0.3724 g of Al(s) reacts with excess hydrochloric acid, we need to use the standard enthalpy of formation and the stoichiometry of the reaction.

The balanced chemical equation for the reaction between Al(s) and hydrochloric acid (HCl) is:

2 Al(s) + 6 HCl(aq) -> 2 AlCl3(aq) + 3 H2(g)

From the balanced equation, we can see that 2 moles of Al(s) react with 6 moles of HCl(aq) to produce 2 moles of AlCl3(aq) and 3 moles of H2(g).

Given that the standard enthalpy of formation of AlCl3(aq) is -538.4 kJ/mol, we can use this information to calculate the heat evolved.

First, we need to convert the mass of Al(s) to moles.

The molar mass of Al is 26.98 g/mol.

Number of moles of Al(s) = mass of Al(s) / molar mass Al
= 0.3724 g / 26.98 g/mol

Next, we need to use the stoichiometry of the reaction to find the moles of AlCl3(aq) produced.

From the balanced equation, we can see that 2 moles of Al(s) react to produce 2 moles of AlCl3(aq).

Therefore, the number of moles of AlCl3(aq) produced = (moles of Al(s) / 2) = (0.3724 g / 26.98 g/mol) / 2

Now, we can calculate the heat evolved using the enthalpy of formation.

Heat evolved = Enthalpy of formation * moles of AlCl3(aq) produced
= -538.4 kJ/mol * [(0.3724 g / 26.98 g/mol) / 2]

Finally, we can substitute the values and calculate the heat evolved.

Heat evolved = -538.4 kJ/mol * [(0.3724 g / 26.98 g/mol) / 2]