Ethane, C2H6 has a molar heat of vaporization of 15 kj/moles. How many kilojoules of energy are required to vaporize 5 grams of Ethane?
Calculate the number of grams in a mole of C2H6, using the chemical formula and the atomic masses of C and H. It will be close to 30 g/mol.
5 g would then be about 5/30 = 1/6 mole.
Multiply the number of moles by the molar heat of vaporization, 15 kJ/mol, for the answer.
2.5
To calculate the amount of energy required to vaporize a given mass of a substance, we need to use the concept of molar heat of vaporization.
First, we need to convert the given mass of ethane from grams to moles. To do this, we need to know the molar mass of ethane.
The molar mass of ethane (C2H6) can be found by adding up the atomic masses of carbon (C) and hydrogen (H). The atomic masses of carbon and hydrogen are 12.01 g/mol and 1.01 g/mol, respectively.
Molar mass of ethane (C2H6) = (2 × atomic mass of carbon) + (6 × atomic mass of hydrogen)
Molar mass of ethane = (2 × 12.01 g/mol) + (6 × 1.01 g/mol)
= 24.02 g/mol + 6.06 g/mol
= 30.08 g/mol
Now, we can calculate the number of moles of ethane in 5 grams using the molar mass.
Number of moles of ethane = Mass of ethane / Molar mass of ethane
= 5 g / 30.08 g/mol
≈ 0.166 moles
Next, we can use the molar heat of vaporization to calculate the energy required to vaporize this many moles of ethane.
Energy required to vaporize ethane = Number of moles × Molar heat of vaporization
= 0.166 moles × 15 kJ/mol
≈ 2.49 kJ
Therefore, approximately 2.49 kilojoules of energy are required to vaporize 5 grams of ethane.