See if this sounds reasonable to you.
moles CaCl2 = 11.11/110.98 = 0.1 mole
Ca = 0.1 mol
Cl = 0.2 mol
Ca = 543 kJ/mol x 0.1 mol = -54.3 kJ.
Cl = 167 kJ/mol x 0.2 mol = -33.4 kJ.
Total released = sum of the two.
deltaHf of Ca 2+ (aq)=-543Kj/mol,
deltaHf of Cl 1-(aq) = -167Kj/mol
CaCl2(s) ==> Ca 2+(aq) + 2Cl 1-(aq)
moles CaCl2 = 11.11/110.98 = 0.1 mole
Ca = 0.1 mol
Cl = 0.2 mol
Ca = 543 kJ/mol x 0.1 mol = -54.3 kJ.
Cl = 167 kJ/mol x 0.2 mol = -33.4 kJ.
Total released = sum of the two.
ΔH_reaction = ΣΔHf(products) - ΣΔHf(reactants)
First, let's convert the mass of calcium chloride (CaCl2) to moles. We can use the molar mass of CaCl2 to do this:
Molar mass of CaCl2 = atomic mass of Ca + (atomic mass of Cl x 2) = 40.08 g/mol + (35.45 g/mol x 2) = 110.98 g/mol
Moles of CaCl2 = mass of CaCl2 / molar mass of CaCl2
= 11.11g / 110.98 g/mol
= 0.1 mol (rounded to one decimal place)
Now, let's calculate the heat of the reaction:
ΔH_reaction = (ΣΔHf(products)) - (ΣΔHf(reactants))
Since CaCl2 is the only reactant, and Ca2+ and 2Cl1- are the products, the equation becomes:
ΔH_reaction = (ΔHf(Ca2+)) + 2(ΔHf(Cl1-)) - ΔHf(CaCl2)
Now, we substitute the given values:
ΔH_reaction = (-543 KJ/mol) + 2(-167 KJ/mol) - 0 KJ/mol
= -543 KJ/mol - 334 KJ/mol
= -877 KJ/mol
Finally, to determine the energy released for the specific mass of calcium chloride, we can use this equation:
Energy released = ΔH_reaction x moles of CaCl2
Energy released = -877 KJ/mol x 0.1 mol
= -87.7 KJ (rounded to one decimal place)
Therefore, the energy released when 11.11g of calcium chloride dissolves in 100 mL of water is approximately -87.7 KJ. The negative sign indicates that the reaction is exothermic, meaning energy is being released.