How much CuO would you need to react with excess aluminum to generate same amout of heat as 139g of Fe2O3 does when reacted with excess aluminum?
2Al + Fe2O3---> Al2O3 + 2Fe delta H=-827kJ
2Al + 3CuO---> Al2O3 + 3Cu delta H=-1170kJ
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Change in enthalpy is an extensive property. This implies that the amount of heat liberated is dependent upon the mass of the reactants. The stoichiometric relationships in respect of mass are
2Al (2x27g) + Fe2O3 (159.7g)---> Al2O3 + 2Fe delta H=-827kJ
2Al (2x27g) + 3CuO (3x79.5)---> Al2O3 + 3Cu delta H=-1170kJ
So, as per the first equation, 139g of Fe2O3, when reacted with sufficient aluminium would generate 139*827/159.7 = 719.8kJ of heat. Clearly, according to stoichiometric relations of the second equation, the same amount of heat shall be generated by 3x79.5x719.8/1170= 146.7g of CuO, when made to react with sufficient amount of aluminium.
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