Liquid Butane, C4H10, is stored in cylinders to be used as a fuel. Suppose 39.3g of butane gas is removed from a cylinder. How much heat must be provided to vaporize this much gas? The heat of vaporization of butane is 21.3 kJ/mol.
answered below.
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To find the amount of heat required to vaporize 39.3g of butane gas, we need to convert the mass of butane to moles and then use the heat of vaporization to calculate the heat energy. Here is how you can do it step by step:
1. Determine the molar mass of butane (C4H10).
- Carbon (C) has a molar mass of 12.01 g/mol.
- Hydrogen (H) has a molar mass of 1.01 g/mol.
- Since there are 4 carbon atoms and 10 hydrogen atoms in butane, the molar mass of butane is:
Molar Mass of Butane = (4 * Molar Mass of Carbon) + (10 * Molar Mass of Hydrogen)
2. Calculate the moles of butane.
- Use the formula: Moles = Mass / Molar Mass
This will give you the number of moles of butane in 39.3g.
3. Apply the stoichiometry of the heat of vaporization to find the heat required.
- The heat of vaporization given is 21.3 kJ/mol.
- Multiply the moles of butane by the heat of vaporization to calculate the heat energy required.
Heat = Moles of Butane * Heat of Vaporization
By following these steps, you can calculate the amount of heat required to vaporize 39.3g of butane gas.