How much heat can be obtained from the condensation of 36.2 g of methanol vapor (mm = 32.04 g/mol) at its boiling point to 36.2 g of liquid methanol at the same temperature? â†hvap = 35.2 kj/mol cliquid = 2.533 j/gâ—¦c cgas = 44.06 j/gâ—¦c

q = -39.8 kJ

Step-by-step explanation:

CH₃OH(g) ⇌ CH₃OH(l); ΔH(cond) = -35.2 kJ·mol⁻¹

(1) Calculate the moles of methanol

n = 36.2 × 1/32.04

n = 1.130 mol

(2) Calculate the heat evolved

q = nΔH(cond)

q = 1.130 × (-35.2)

q = -39.8 kJ

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