q = n * ∆H_vap
Where q is the amount of energy required, n is the number of moles of the substance being boiled, and ∆H_vap is the heat of vaporization.
First, we need to calculate the mass of 2.78 moles of iron:
m = n * M
Where M is the molar mass of iron, which is approximately 55.85 g/mol.
m = 2.78 moles * 55.85 g/mol m = 155.3 g
Next, we can use the following formula to calculate the amount of energy required to raise the temperature of 155.3 g of iron from room temperature to its boiling point of 2800 °C:
q1 = m * Cp * ∆T
Where Cp is the specific heat capacity of iron and ∆T is the change in temperature.
The specific heat capacity of iron is approximately 0.45 J/g°C.
∆T = 2800 °C - 25 °C ∆T = 2775 °C
q1 = 155.3 g * 0.45 J/g°C * 2775 °C q1 = 191,192.25 J
Finally, we can use the formula above to calculate the amount of energy required to vaporize 155.3 g of iron:
q2 = n * ∆H_vap
q2 = 2.78 moles * 350,000 J/mol q2 = 973,000 J
The total amount of energy required to boil 2.78 moles of iron is the sum of q1 and q2:
q = q1 + q2 q = 191,192.25 J + 973,000 J q = 1,164,192.25 J
Therefore, approximately 1,164,192.25 J of energy is required to boil 2.78 moles of iron at its boiling point of 2800 °C.