Answer:
The enthalpy if 68.10 grams of CO2 is produced is -189.04 kJ
Explanation:
<u>Step 1:</u> Data given
temperature = 850 °C
Mass of 68.10 grams of CO2
ΔH°f (CaO) = -635.6 kJ/mol
ΔH°f (CO2) = -693.5 kJ/mol
ΔH°f (CaCO3) =-1206.9 kJ/mol
<u>Step 2: </u>The balanced equation
CaCO3(s) → CaO(s) + CO2(g)
<u>Step 3: </u>Calculate ΔH°reaction
ΔH°reaction = ΣΔH°f (products) - ΣΔH°f (reactants)
ΔH°reaction = (ΔH°f (CaO) + ΔH°f (CO2)) - ΔH°f (CaCO3)
ΔH°reaction = (-635.6 kJ/mol + -693.5 kJ/mol) + 1206.9 kJ/mol
ΔH°reaction = -122.2 kJ /mol
<u>Step 4:</u> Calculate moles of CO2
Moles CO2 = mass CO2 / Molar mass CO2
Moles CO2 = 68.10 grams / 44.01 g/mol
Moles CO2 = 1.547 moles
<u>Step 5:</u> Calculate the enthalpy change for 68.10 grams of CO2
-122.2 kJ/mol * 1.547 moles = -189.04 kJ
The enthalpy if 68.10 grams of CO2 is produced is -189.04 kJ
