Ethanol (C2H5OH) is synthesized for industrial use by the following reaction, carried out at very high pressure.
C2H4(g) + H2O(g) C2H5OH(l)
What is the maximum amount of ethanol (in grams) that can be produced when 2.2 kg of ethylene (C2H4) and 0.014 kg of steam are placed into the reaction vessel?
_______________g
See your other post above.
To determine the maximum amount of ethanol that can be produced, we need to do the following steps:
Step 1: Convert the given masses of ethylene (C2H4) and steam into moles.
Step 2: Identify the limiting reagent based on the stoichiometry of the reaction.
Step 3: Calculate the amount of ethanol produced from the limiting reagent.
Step 4: Convert the moles of ethanol obtained in Step 3 into grams.
Step 1:
The molar mass of ethylene (C2H4) is calculated as follows:
C: 12.01 g/mol x 2 = 24.02 g/mol
H: 1.01 g/mol x 4 = 4.04 g/mol
Total molar mass of C2H4 = 24.02 g/mol + 4.04 g/mol = 28.06 g/mol
To convert the given mass of ethylene (2.2 kg) into moles, we can use the following equation:
moles = mass / molar mass
moles of C2H4 = 2.2 kg / 28.06 g/mol = 78.49 mol
The molar mass of steam (H2O) is:
H: 1.01 g/mol x 2 = 2.02 g/mol
O: 16.00 g/mol
Total molar mass of H2O = 2.02 g/mol + 16.00 g/mol = 18.02 g/mol
To convert the given mass of steam (0.014 kg) into moles:
moles = mass / molar mass
moles of H2O = 0.014 kg / 18.02 g/mol = 0.777 mol
Step 2:
Using the balanced equation, we can see that for every 1 mole of ethylene (C2H4), we need 1 mole of steam (H2O), and we will produce 1 mole of ethanol (C2H5OH).
The moles of ethylene (C2H4) and steam (H2O) are equal, so neither is in excess or limiting in this case.
Step 3:
Since the moles of ethylene and steam are equal, we can calculate the moles of ethanol generated using either the moles of C2H4 or H2O. Let's use the moles of C2H4.
moles of ethanol = moles of C2H4 = 78.49 mol
Step 4:
To convert the moles of ethanol into grams, we need to multiply by the molar mass of ethanol.
The molar mass of ethanol (C2H5OH) 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 + 1.01 g/mol = 17.01 g/mol
Total molar mass of C2H5OH = 24.02 g/mol + 6.06 g/mol + 17.01 g/mol = 47.09 g/mol
grams of ethanol = moles of ethanol x molar mass of ethanol
grams of ethanol = 78.49 mol x 47.09 g/mol = 3,695.15 g
Therefore, the maximum amount of ethanol that can be produced is 3,695.15 grams.