A student determines the value of the equilibrium constant to be 4.67×1091 for the following reaction. H2CO(g) + O2(g)CO2(g) + H2O(l) Based on this value of Keq: G° for this reaction is expected to be (greater, less) fill in the blank 1 than zero. Calculate the free energy change for the reaction of 2.16 moles of H2CO(g) at standard conditions at 298K.

kuchisake

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explanation:

"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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the answer to this question is 3.