Determine the standard enthalpy of formation of SrI2(s) using a Born–Haber cycle. Enthalpy of sublimation of Sr(s) = 164 kJ/mol 1st ionization energy of Sr(g) = 549 kJ/mol 2nd ionization energy of Sr(g) = 1064 kJ/mol Enthalpy of sublimation of I2(s) = 62.4 kJ/mol Bond dissociation energy of I2(g) = 152.55 kJ/mol 1st electron affinity of I(g) = –295.15 kJ/molLattice energy of SrI2(s) = –1959.75 kJ/mol

Enthalpy of formation of strontium chloride, SrCl = -558.1 kJ/mol

Explanation:

Sr(s) > Sr(g)       ΔHsublimation = +164 kJ/mol

Sr⁺(g) > Sr⁺(g)    First Ionization energy, IE₁ = +549 kJ/mol

Sr⁺(g) > Sr²⁺(g)    Second ionization energy, IE₂ = +1064 kJ/mol

I₂(s) > I₂(g)   ΔHsublimation = +62.4 kJ/mol

I₂(g) > 2I(g)   Bond dissociation energy, BE = +152.55 kJ/mol

2I(g) > 2I⁻(g)   Electron affinity, EA = 2 (-295.15 kJ/mol) = -590.3 kJ/mol

Sr²⁺(g) + 2I⁻(g) > SrI₂(s)    Lattice energy, LE = -1959.75 kJ/mol

Overall equation for the formation of SrCl₂ is given as:

Sr(s) + I₂(s) > SrI₂(s)    ΔHformation = ?

Enthalpy of formation, ΔHformation = (ΔHsublimation of Sr(s) + IE₁ + IE₂ + ΔHsublimation of I₂(s) + BE + EA + LE)

ΔHformation = {164 + 549 + 1064 + 62.4 + 152.55 + (-590.3) + (-1959.75)} kJ/mol

Sr(s) + I₂(s) > SrI₂(s)    ΔHformation = -558 kJ/mol

Therefore, enthalpy of formation of strontium chloride, SrCl = -558.1 kJ/mol

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)

"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"