Question

Al2(SO4)3(s) + H2O(l) ---- Al2O3(s) + H2SO4 (aq) calculate enthalpy formation for this reaction. Balance the reaction. Calculate the total enthalpy change that would occur from 15 moles of Al2(SO4)3

Answer 1

The given chemical equation is:

$Al_{2}(SO_{4})_{3}(aq)+H_{2}O(l)-->Al_{2}O_{3}(aq)+H_{2}SO_{4}(aq)$

On balancing the equation we get,

$Al_{2}(SO_{4})_{3}(aq)+3H_{2}O(l)-->Al_{2}O_{3}(aq)+3H_{2}SO_{4}(aq)$

Calculating enthalpy of formation of this reaction from the standard heats of formation of the products and reactants:

Δ$H_{reaction}^{0}=[H_{f}^{0}(Al_{2}O_{3}(s)) + (3*H_{f}^{0}(H_{2}SO_{4}(aq))] - [H_{f}^{0}(Al_{2}SO_{4}(aq)) + (3*H_{f}^{0}(H_{2}O(l))]$

=[(-1669.8kJ/mol)+ {3* (-909.27 kJ/mol)}]-[(-3442kJ/mol)+{3*(-285.8 kJ/mol)}]

=[(-4397.61kJ/mol)]-[(-4299.4kJ/mol)]

=-98.21kJ/mol

Total enthalpy change when 15 mol of $Al_{2}(SO_{4})_{3}$reacts will be=

$15 mol Al_{2}(SO_{4})_{3}*\frac{-98.21kJ}{1 molAl_{2}(SO_{4})_{3}} =-1473.15kJ/mol$