Question

Use the following information on Al to determine the amounts of heat for the three heating steps required to convert 150.0 g of solid Al at 458oC into liquid Al at 758oC. mp = 658 oC bp = 2467 o C
Molar Heat Capacities
Csolid = 24.3 J/moloC Cliquid = 29.3 J/mol oC
ΔHfusion = 10.6 kJ/mol ΔHvaporization = 284 kJ/mol

Answer 1

Al moles are calculated:

n Al = g / MM = 150 g / 27 g / mol = 5.6 mol

The heat required at each stage is calculated:

i) Heat solid up to mp:

qi = n * cs * ΔT = 5.6 mol * 0.0243 kJ / mol * ° C * (658 - 458) ° C = 27.22 kJ

ii) Melt Al:

qii = ΔHf * n = 10.6 kJ / mol * 5.6 mol = 59.36 kJ

iii) Heat to liquid:

qiii = 5.6 mol * 0.0293 kJ / mol * ° C * (758 - 658) ° C = 16.41 kJ

The total heat is:

q total = qi + qii + qiii = 102.99 kJ

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