When alcohol evaporates according to the equationC2H5OH(l) --> C2H5OH(g), how many liters of gaseous alcohol, measured at STP, are created by the evaporation of 566.1 grams of alcohol?

To calculate the number of liters of gaseous alcohol produced by the evaporation of 566.1 grams of alcohol, we need to use the concept of molar mass and the ideal gas law.

Here's how you can solve it step by step:

Step 1: Calculate the number of moles of alcohol.
The molar mass of C2H5OH (ethyl alcohol) is:
C = 12.01 g/mol x 2 = 24.02 g/mol
H = 1.01 g/mol x 6 = 6.06 g/mol
O = 16.00 g/mol x 1 = 16.00 g/mol

Add all the atomic masses together:
24.02 g/mol + 6.06 g/mol + 16.00 g/mol = 46.08 g/mol

Now, divide the mass of alcohol (566.1 grams) by the molar mass to obtain moles:
566.1 g / 46.08 g/mol = 12.28 moles

Step 2: Apply the ideal gas law.
The ideal gas law is given by:
PV = nRT

Where:
P = pressure (measured in atm)
V = volume (measured in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (measured in Kelvin)

At STP (Standard Temperature and Pressure), the temperature is 273 K, and the pressure is 1 atm.

Since we are calculating the volume, we can rearrange the ideal gas law to solve for V:
V = nRT / P

Step 3: Substitute the known values into the equation.
V = (12.28 moles) * (0.0821 L·atm/mol·K) * (273 K) / (1 atm)

Step 4: Calculate the volume of gaseous alcohol.
V = 28.30 L

Therefore, the evaporation of 566.1 grams of alcohol would create approximately 28.30 liters of gaseous alcohol when measured at STP.