Calculating Heat Change: 15.8 g CH3OH Condensation at 25C (ΔHvap=38.0 kJ/mol)

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SUMMARY

The discussion focuses on calculating the heat released during the condensation of 15.8 g of methanol (CH3OH) at 25°C, utilizing the heat of vaporization (ΔHvap) of 38.0 kJ/mol. To determine the heat released, participants emphasize the need to convert grams of CH3OH to moles using its molar mass, which is approximately 32.04 g/mol. The calculation involves using the formula Q = n × ΔHvap, where Q is the heat released, n is the number of moles, and ΔHvap is the heat of vaporization. The final result indicates that 15.8 g of CH3OH releases approximately 19.0 kJ of heat upon condensation.

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hey guys, I'm little stuck on this question: How much heat, in kJ, is released when 15.8 g CH3OH is condensed at 25C? (ΔHvap=38.0 kJ/mol)

I know that the equation is: H= mΔT x SH, for change in temp.
and the phase change: H=mΔV

I have no idea how to set it up according to the given variables.
 
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Assuming the condensation occurs at a constant temperature you need your second equation only.You need 38kJ for one mole so you need to find out how many moles there are in 15.8g
 

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