When 25.0 g of ethanol vapor condenses, 23.6 kJ of heat is released. How much heat is required in kJ to vaporize 254 g of ethanol?
23.6 kJ x (254 g/25.0 g)= ??
To find out how much heat is required to vaporize 254 g of ethanol, we need to use the concept of molar heat of vaporization.
1. First, we need to calculate the moles of ethanol vapor condensed in the first scenario. To do this, we will use the molar mass of ethanol, which is 46.07 g/mol.
Moles of ethanol vapor condensed = mass / molar mass
= 25.0 g / 46.07 g/mol
2. Next, we can use the amount of heat released during the condensation process (23.6 kJ) to calculate the molar heat of vaporization (ΔHvap).
ΔHvap = heat released / moles of ethanol vapor condensed
= 23.6 kJ / (25.0 g / 46.07 g/mol)
3. Now we can use the molar heat of vaporization to calculate the amount of heat required to vaporize 254 g of ethanol.
Heat required to vaporize 254 g of ethanol = moles of ethanol / ΔHvap
= 254 g / 46.07 g/mol * ΔHvap
Note: Since we don't have the exact value of ΔHvap, we can only provide the general method to calculate it. If you have the specific value of ΔHvap for ethanol, you can substitute it into the equation to get the precise heat required.
To find the amount of heat required to vaporize 254 g of ethanol, we can use the concept of heat of vaporization. The heat of vaporization is the amount of heat energy required to convert a liquid into a gas at a constant temperature and pressure.
Given that 25.0 g of ethanol releases 23.6 kJ of heat when condensing, we can first find the heat of vaporization for ethanol using the given data.
Heat of vaporization (ΔHvap) = Heat released / Mass of substance
ΔHvap = 23.6 kJ / 25.0 g
Now we can use this calculated value of heat of vaporization to find the amount of heat required to vaporize 254 g of ethanol.
Heat required = ΔHvap * Mass of substance
Heat required = ΔHvap * 254 g
Substituting the calculated value of ΔHvap:
Heat required = (23.6 kJ / 25.0 g) * 254 g
Now we can perform the calculation to find the answer.
Heat required = 23.6 kJ / 25.0 * 254 g = 240.808 kJ
Therefore, the amount of heat required to vaporize 254 g of ethanol is approximately 240.808 kJ.