For a given reaction, δh = -19.9 kj/mol and δs = -55.5 j/kmol.

the reaction will have δg = 0 at k.

assume that δh and δs do not vary with temperature.

359 K

Explanation:

The Gibbs free energy can be calculated using the following expression.

ΔG = ΔH - T.ΔS

where,

ΔG is the Gibbs free energy

ΔH is the enthalpy of the reaction

T is the absolute temperature

ΔS is the entropy of the reaction.

When ΔG = 0,

ΔH - T.ΔS = 0

ΔH = T.ΔS

T = ΔH/ΔS = (-19.9 × 10³ J/mol) / (-55.5 J/K.mol) = 359 K

"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"