Some properties of aluminum are summarized in the following list.

normal melting point 658°C
heat of fusion 0.395 kJ/g
normal boiling point 2467°C
heat of vaporization 10.52 kJ/g
specific heat of the solid 0.902 J/g°C

(a) Calculate the quantity of energy required to heat 1.75 mol of aluminum from 17°C to its normal melting point.

(b) Calculate the quantity of energy required to melt 1.37 mol of aluminum at 658°C.

(c) Calculate the amount of energy required to vaporize 1.36 mol of aluminum at 2467°C.

a)

q = mass x specific heat x (Tfinal-Tinitial).

b)
q = mass x heat fusion.

c)
q = mass x heat vap

This is how I'm doing part A. I'm told it's the wrong answer, but I can't figure out what I'm doing wrong. I'm hesitant on going to parts B and C until I understand part A.

Part A:

Step one: I changed the mols of Al to grams.

13.9mol Al * (26.9815g Al / 1mol Al) = 47.2176g Al

Step two: I plugged the numbers into the Q = s * m * change in T

Q = (0.902 J/g*C)(47.2176g Al) ( 641C)
Q = 27300KJ = 2.73 * 10^4KJ

So the answer to part A is 2.73*10^4kJ

Again, I'm told I'm wrong here. Any help would be appreciated thanks!

Forget it... I got it thanks!!!!!

I answered this below where you showed your work. I think the problem is that you have kJ and it is actually J.

To calculate the quantity of energy required for heating, melting, and vaporizing aluminum, we need to use the formulas that relate the energy to the specific heat, heat of fusion, and heat of vaporization.

(a) Quantity of energy required to heat aluminum from 17°C to its normal melting point:
First, we need to calculate the energy required to heat the aluminum from 17°C to its melting point. We can use the formula:
Q = mcΔT

Where:
Q = energy (in joules)
m = mass (in grams)
c = specific heat (in J/g°C)
ΔT = change in temperature (in °C)

Since we have the specific heat of the solid aluminum, we can directly substitute the given values into the formula:
m = 1.75 mol of aluminum × molar mass of aluminum
molar mass of aluminum = 26.98 g/mol

Now, let's calculate the mass of aluminum:
m = 1.75 mol × 26.98 g/mol = 47.17 g

ΔT = normal melting point - initial temperature
ΔT = 658°C - 17°C = 641°C

Substituting the values into the formula:
Q = (47.17 g) × (0.902 J/g°C) × (641°C)

Calculate the product of the three values in the parentheses, then multiply it by the mass of aluminum:
Q = (47.17 g) × (0.902 J/g°C) × (641°C) = 2.166 × 10^4 J

The quantity of energy required to heat 1.75 mol of aluminum from 17°C to its normal melting point is 2.166 × 10^4 Joules.

(b) Quantity of energy required to melt aluminum:
To calculate the energy required to melt aluminum, we use the heat of fusion formula:

Q = n × ΔHf

Where:
Q = energy (in joules)
n = number of moles (in mol)
ΔHf = heat of fusion (in J/mol)

Substituting in the given values:
n = 1.37 mol
ΔHf = 0.395 kJ/g × 26.98 g/mol (since we need to convert kJ/g to J/mol)

Now, let's calculate the heat of fusion:
ΔHf = 0.395 kJ/g × 26.98 g/mol = 10.65 kJ/mol = 10.65 × 10^3 J/mol

Substituting the values into the formula:
Q = (1.37 mol) × (10.65 × 10^3 J/mol) = 1.457 × 10^4 J

The quantity of energy required to melt 1.37 mol of aluminum at 658°C is 1.457 × 10^4 Joules.

(c) Quantity of energy required to vaporize aluminum:
To calculate the energy required to vaporize aluminum, we use the heat of vaporization formula:

Q = n × ΔHv

Where:
Q = energy (in joules)
n = number of moles (in mol)
ΔHv = heat of vaporization (in J/mol)

Substituting in the given values:
n = 1.36 mol
ΔHv = 10.52 kJ/g × 26.98 g/mol (since we need to convert kJ/g to J/mol)

Now, let's calculate the heat of vaporization:
ΔHv = 10.52 kJ/g × 26.98 g/mol = 28.37 kJ/mol = 28.37 × 10^3 J/mol

Substituting the values into the formula:
Q = (1.36 mol) × (28.37 × 10^3 J/mol) = 3.86 × 10^4 J

The quantity of energy required to vaporize 1.36 mol of aluminum at 2467°C is 3.86 × 10^4 Joules.