Question

Calculate the amount of heat released in the complete combustion of 8.17 grams of Al to form Al2O3(s) at 25°C and 1 atm. for Al2O3(s) = −1676 kJ/mol 4Al(s) + 3O2(g) → 2Al2O3(s)

Answer 1

Answer – Given, mass of Al = 8.17 g , ∆H of Al₂O₃(s) = −1676 kJ/mol

4Al(s) + 3O₂(g) -----> 2Al₂O₃(s)

First we need to calculate the moles of Al

Moles of Al = 8.17 g / 26.982 g.mol^-1

                    = 0.303 moles

From the above balanced reaction

3 moles of Al = 2 moles of Al₂O₃(s)

So, 0.303 moles of Al = ?

= 0.202 moles of Al₂O₃(s)

We know,

1 mole of Al₂O₃(s) = -1676 kJ

So, 0.202 moles of Al₂O₃(s) = ?

= -338.3 kJ

The amount of heat released in the complete combustion of 8.17 grams of Al to form Al₂O₃(s) at 25°C and 1 atm is -338.3 kJ.