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?
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.
To solve this problem, we need to use the concept of stoichiometry and the molar heat of vaporization of ethanol.
First, we need to convert the given heat energy from kilojoules (kJ) to joules (J). We can do this by multiplying the value by 1000 (1 kJ = 1000 J):
343.0 kJ * 1000 = 343000 J
Next, we need to determine the number of moles of ethanol required to absorb this amount of energy. We can do this by using the molar heat of vaporization of ethanol:
38.6 kJ/mol
To find the number of moles, we can divide the heat energy in joules by the molar heat of vaporization of ethanol:
Number of moles = 343000 J / 38.6 kJ/mol
Now, we need to convert the number of moles of ethanol to grams. To do this, we need to know the molar mass of ethanol:
C: 12.01 g/mol
H: 1.01 g/mol
O: 16.00 g/mol
The molar mass of ethanol (C2H5OH) is:
2(12.01 g/mol) + 6(1.01 g/mol) + 1(16.00 g/mol) = 46.07 g/mol
Finally, we can calculate the mass of ethanol in grams by multiplying the number of moles of ethanol by its molar mass:
Mass of ethanol = Number of moles * Molar mass of ethanol
Remember to multiply the number of moles by the molar mass of ethanol:
Number of moles = 343000 J / (38.6 kJ/mol) = 8885.02 mol
Mass of ethanol = 8885.02 mol * 46.07 g/mol
Thus, the amount of ethanol that can be boiled with 343.0 kJ of heat energy is approximately 409,249 grams.
To solve this problem, we can use the equation:
q = n * ΔH
Where:
q is the amount of heat energy absorbed or released (343.0 kJ in this case)
n is the number of moles of ethanol
ΔH is the molar heat of vaporization of ethanol (38.6 kJ/mol)
To find the number of moles, we can rearrange the equation as:
n = q / ΔH
Plugging in the values, we get:
n = 343.0 kJ / 38.6 kJ/mol
To find the number of grams, we need to multiply the number of moles by the molar mass of ethanol, which is 46.08 g/mol.
n (in moles) * molar mass (in grams/mol) = mass (in grams)
Let's calculate:
n = 343.0 kJ / 38.6 kJ/mol = 8.89 mol
mass = 8.89 mol * 46.08 g/mol = 409.5 g
Therefore, 343.0 kJ of heat energy can boil approximately 409.5 grams of ethanol.