How much energy will be released when 11.11g of calcium chloride dissolves in 100 mL of water.
deltaHf of Ca 2+ (aq)=-543Kj/mol,
deltaHf of Cl 1-(aq) = -167Kj/mol
CaCl2(s) ==> Ca 2+(aq) + 2Cl 1-(aq)
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.
To determine the energy released when 11.11g of calcium chloride dissolves in 100 mL of water, we need to calculate the heat of the reaction. The heat of the reaction can be calculated using the following formula:
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.