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, in kg, of ethanol that can be produced when 1.65 kg of ethylene (C2H4) and 0.0610 kg of steam are placed into the reaction vessel?
Get the Limiting Reactant of your reaction,
so get the Molar Mass (MM) of C2H4 and H2O from your periodic table
then
1650 g / MM of C2H4 = __ mol
61 g / MM of H2O = __ mol
the least amount of mol will be your Limiting Reactant.
You use the LR to find the max amount of C2H5OH you will get from your reaction
so
LR in mol * MM of C2H5OH = ___ g
divide answer to get your max amt. in kg
*divide answer by 1000 to get max amt in kg
thanks!
To determine the maximum amount of ethanol that can be produced, we need to use the concept of limiting reactants.
1. Start by calculating the number of moles for each reactant:
- Ethylene (C2H4):
1.65 kg * (1 mol/28.05 g) = 0.0587 mol
- Steam (H2O):
0.0610 kg * (1 mol/18.02 g) = 0.0034 mol
2. Examine the stoichiometry of the balanced chemical equation to determine the mole ratio between the reactants and the product:
C2H4(g) + H2O(g) → C2H5OH(l)
1 mol C2H4 : 1 mol H2O : 1 mol C2H5OH
3. Identify the limiting reactant by comparing the mole ratios of the reactants:
Since the mole ratio between C2H4 and H2O is 1:1, it appears that C2H4 is the limiting reactant as it is present in a greater quantity.
4. Calculate the maximum number of moles of ethanol that can be produced using the moles of limiting reactant:
0.0587 mol C2H4 * (1 mol C2H5OH/1 mol C2H4) = 0.0587 mol C2H5OH
5. Convert the moles of ethanol to kilograms using its molar mass:
0.0587 mol C2H5OH * 46.07 g/mol = 2.705 g
6. Finally, convert grams to kilograms:
2.705 g * (1 kg/1000 g) = 0.00271 kg
Therefore, the maximum amount of ethanol that can be produced is 0.00271 kg.