Calculate the enthalpy change for the thermite reaction: 2al(s)+fe2o3(s)→2fe(s)+al2o3(s), δh∘rxn=−850 kj when 12.0 mol of al undergoes the reaction with a stoichiometrically equivalent amount of fe2o3.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

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