Calculate the AH for the following balanced reactions. rxn
→
Remember the AH is for the reaction written.
the
1) Fe, O3 (5) + 2 AL 6 ALO3(s) + 2 Fe (3)
2) 2 C, H6 + 7 02 4 CO, + 6 H2O

"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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