Question

Calculate the minimum amount (in grams) of O2 needed to produce 243.2 kJ of heat, when it is reacted with Al(s) to form Al2O3(s) at 25°C and 1 atm. Delta Δ Hf Al2O3(s) = − 1676 kJ/mol

Answer 1

The balanced equation can be written as 4Al(s) + 3O₂(g) $\rightarrow$ 2Al₂O₃(s)

The amount of heat produced per 1 mole of Al2O3(s) = 1676 kJ

Therefore, the no. of moles of Al2O3(s) formed to produce 243.2 KJ heat = 243.2/1676 = 0.1451074 mol

For the formation of 2 moles of Al2O3(s), the no. of moles of O2(g) needed = 3

i.e. For the formation of 0.1451 mol Al2O3(s), the no. of moles of O2(g) needed = 0.1451074*3/2 = 0.2176611 mol

Therefore, the minimum amount of O2 needed = 0.2176611 mol * 32 g/mol = 6.965 g