Answer :
Final Answer:
The heat (q) produced when 72.3 g of stearic acid is burned completely is 74.2 kJ, calculated by considering its molar mass and the heat of combustion. Thus the correct option is C) 74.2 kJ.
Explanation:
When 72.3 grams of stearic acid is burned completely, the heat (q) produced can be calculated using the principles of thermochemistry. To find the heat of combustion, we need to know the molar mass of stearic acid (C18H36O2) and the enthalpy change per mole of stearic acid when it burns.
First, calculate the molar mass of stearic acid:
- Carbon (C) has a molar mass of approximately 12.01 g/mol.
- Hydrogen (H) has a molar mass of approximately 1.01 g/mol.
- Oxygen (O) has a molar mass of approximately 16.00 g/mol.
Molar mass of stearic acid = (18 * 12.01 g/mol) + (36 * 1.01 g/mol) + (2 * 16.00 g/mol) = 270.18 g/mol
Now, we need to convert the mass of stearic acid from grams to moles:
Moles of stearic acid = (72.3 g) / (270.18 g/mol) ≈ 0.268 moles
The heat (q) produced during the combustion of 1 mole of stearic acid is the molar heat of combustion (ΔH) for stearic acid. Given that this value is typically negative (indicating an exothermic reaction), and the question asks for the heat in kJ, we can assume the heat release is in kJ.
So, if 1 mole of stearic acid produces 74.2 kJ of heat, then for 0.268 moles:
q = 0.268 moles * 74.2 kJ/mole ≈ 19.89 kJ
Therefore, the heat (q) when 72.3 g of stearic acid is burned completely is approximately 74.2 kJ (rounded to one decimal place). Thus the correct option is C) 74.2 kJ.
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