Asked by Rose
Calculate the enthalpy of formation of maganese (IV) oxide based on the following information.
4Al(s) + 3MnO2(s) --> 3Mn(s) + 2Al2O3(s) Delta H = -1790 kJ
2Al(s) + 3/2O2 (g) ---> Al2O3 (s) Delta Hf = -1676 kJ
4Al(s) + 3MnO2(s) --> 3Mn(s) + 2Al2O3(s) Delta H = -1790 kJ
2Al(s) + 3/2O2 (g) ---> Al2O3 (s) Delta Hf = -1676 kJ
Answers
Answered by
bobpursley
we need twice the second reaction to make 2Al2O3, then the first reaction reversed to make three moles of MnO2
Hr=2*(-1676kJ)+1790kJ
but we make 3 moles MnO2
hf=Hr above/3 kJ/mole
Hr=2*(-1676kJ)+1790kJ
but we make 3 moles MnO2
hf=Hr above/3 kJ/mole
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