How many grams of ethanol, C2H5OH, can be boiled with 343.0 kJ of heat energy? The molar heat of vaporization of ethanol is 38.6 kJ/mol

I'm not sure how I am supposed to set this problem up. Can someone help me, please?

1 answer

Sure! To solve this problem, you can use the following equation:

q = n * ΔHvap

where:
q = heat energy (in kJ)
n = number of moles
ΔHvap = molar heat of vaporization (in kJ/mol)

First, we need to determine the number of moles of ethanol that can be boiled with the given heat energy. We can rearrange the equation and solve for n:

n = q / ΔHvap

Substituting the given values:

n = 343.0 kJ / 38.6 kJ/mol
n ≈ 8.88 moles

Now that we have the number of moles, we can convert this to grams using the molar mass of ethanol:

m = n * M

where:
m = mass of ethanol (in grams)
M = molar mass of ethanol (in g/mol)

The molar mass of ethanol is 46.07 g/mol (2 * 12.01 g/mol for carbon + 6 * 1.01 g/mol for hydrogen + 1 * 16.00 g/mol for oxygen). Now we can calculate the mass of ethanol:

m = 8.88 moles * 46.07 g/mol
m ≈ 409.1 g

So, approximately 409.1 grams of ethanol can be boiled with 343.0 kJ of heat energy.