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How much heat (in kilocalories) is released when 8.2 mol of butane freezes at -138.4°C?

A. 38.4 kilocalories
B. 76.8 kilocalories
C. 15.36 kilocalories
D. 30.72 kilocalories

Answer :

Final answer:

The amount of heat released when 8.2 mol of butane freezes can be calculated using the equation q = m * ΔH. The correct answer is none of the options given in the question.

Explanation:

When a substance freezes, heat is released. The amount of heat released can be calculated using the equation q = m * ΔH, where q is the heat released, m is the mass of the substance, and ΔH is the enthalpy change per mole of the substance during freezing. Given that butane has a molar mass of 58.12 g/mol and its enthalpy change of fusion is -125.7 kJ/mol, we can calculate the amount of heat released when 8.2 mol of butane freezes:

q = 8.2 mol * -125.7 kJ/mol

q = -125.7 kJ/mol * 8.2 mol

q = -1030.14 kJ

Converting kJ to kilocalories:

1 kcal = 4.184 kJ

-1030.14 kJ * (1 kcal / 4.184 kJ) = -246.42 kcal

The amount of heat released when 8.2 mol of butane freezes is approximately 246.42 kilocalories. Therefore, none of the given options a), b), c), or d) are correct.

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