If 448 L of ethene gas are reacted with excess water at STP, how many grams of ethanol will be produced?
To determine the number of grams of ethanol produced when reacting 448 L of ethene gas with excess water at STP (Standard Temperature and Pressure), we need to use the balanced chemical equation for the reaction and convert between volume and mass using the ideal gas law and molar mass.
The balanced chemical equation for the reaction is:
C2H4 (ethene) + H2O (water) → C2H5OH (ethanol)
Step 1: Convert the given volume of ethene gas from liters to moles.
To do this, we need to use the ideal gas law, which states: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
At STP:
- Pressure (P) = 1 atm
- Temperature (T) = 273 K
Using the value of the ideal gas constant (R) = 0.0821 L.atm/mol.K, we can calculate the number of moles (n).
n = PV / RT = (1 atm) * (448 L) / (0.0821 L.atm/mol.K * 273 K)
Step 2: Determine the molar ratio between ethene and ethanol.
From the balanced chemical equation, we can see that the ratio between C2H4 and C2H5OH is 1:1.
Step 3: Convert moles of ethene to moles of ethanol.
Since the moles of ethene and ethanol are the same according to the balanced equation, the number of moles of ethanol produced is equal to the number of moles of ethene calculated in step 1.
Step 4: Convert moles of ethanol to grams.
To convert moles to grams, we need to use the molar mass of ethanol, which is 46.07 g/mol.
Mass of ethanol = Number of moles of ethanol * Molar mass of ethanol
By using this method, you can determine the number of grams of ethanol that will be produced when 448 L of ethene gas reacts with excess water at STP.