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.
39.3 g butane* 21.3 kJ/mol = ??,
Convert 39.3 g butane to mols and use the above equation.
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To find the heat required to vaporize the given amount of butane gas, we need to follow these steps:
Step 1: Calculate the number of moles of butane gas.
To do this, we can use the molar mass of butane, which is 58.12 g/mol.
The number of moles (n) can be calculated using the formula:
n = mass / molar mass
Given mass = 39.3 g,
n = 39.3 g / 58.12 g/mol
Step 2: Calculate the heat required to vaporize the butane gas.
The heat required (q) can be calculated using the formula:
q = n * ΔHvap
Given ΔHvap (enthalpy of vaporization) = 21.3 kJ/mol
Note that we need to convert kJ to J by multiplying by 1000 to match the units.
q = n * ΔHvap * 1000
Now, let's substitute the values and calculate the result.
n = 39.3 g / 58.12 g/mol ≈ 0.677 mol
q = 0.677 mol * 21.3 kJ/mol * 1000 = 14,432.1 J
Therefore, approximately 14,432.1 J (or 14.4 kJ) of heat must be provided to vaporize 39.3 g of butane gas.
To calculate the amount of heat required to vaporize a given amount of butane gas, we need to follow these steps:
Step 1: Convert grams of butane gas to moles.
The molar mass of butane (C4H10) can be calculated by adding the molar masses of carbon and hydrogen.
Carbon (C) has a molar mass of 12.01 g/mol, and there are four carbon atoms in butane, so the carbon contribution to the molar mass is 12.01 g/mol x 4 = 48.04 g/mol.
Hydrogen (H) has a molar mass of 1.01 g/mol, and there are ten hydrogen atoms in butane, so the hydrogen contribution to the molar mass is 1.01 g/mol x 10 = 10.10 g/mol.
Adding these two contributions together gives us the molar mass of butane:
48.04 g/mol + 10.10 g/mol = 58.14 g/mol.
To convert grams of butane gas to moles, we need to divide the mass by the molar mass:
39.3 g / 58.14 g/mol = 0.676 moles.
Step 2: Calculate the heat of vaporization.
Given that the heat of vaporization of butane is 21.3 kJ/mol, we can use this value to calculate the heat required to vaporize the amount of butane gas we found in step 1.
0.676 moles x 21.3 kJ/mol = 14.4078 kJ.
Therefore, approximately 14.4 kJ of heat must be provided to vaporize 39.3 g of butane gas.