What is the maximum amount of ethanol (in grams) that can be produced when 0.30 kg of ethylene (C2H4) and 0.029 kg of steam are placed into the reaction vessel?
Do you have an equation? If so then this is a limiting reagent problem. I solve these problems the long way by solving two stoichiometry problems. Here is a worked example of a stoichiometry problem. Just follow the steps AND the directions below the link.
http://www.jiskha.com/science/chemistry/stoichiometry.html
First use the amount of ethylene listed, assume you have all of the steam you need, and calculate, following the example problem, the moles ethanol produced. Then use the grams steam, assume you have all of the ethylene you need, and follow the example to calculate the moles ethanol. You will obtain two answers; both can't be right. The correct answer, in limiting reagent problems, is ALWAYS the smaller value. Using that value, follow the example to find grams ethanol.
To find the maximum amount of ethanol that can be produced, we first need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To determine the limiting reagent, we need to compare the amounts of ethylene and steam (water) in terms of moles. Let's calculate the moles of each reactant.
1. Calculate the moles of ethylene (C2H4):
- The molar mass of ethylene (C2H4) = 2 * atomic mass of carbon (C) + 4 * atomic mass of hydrogen (H)
= (2 * 12.01 g/mol) + (4 * 1.01 g/mol)
= 28.05 g/mol
- Moles of ethylene = Mass of ethylene / Molar mass of ethylene
= 0.30 kg / 28.05 g/mol
= 10.70 mol
2. Calculate the moles of steam (H2O):
- The molar mass of water (H2O) = 2 * atomic mass of hydrogen (H) + atomic mass of oxygen (O)
= (2 * 1.01 g/mol) + 16.00 g/mol
= 18.02 g/mol
- Moles of water = Mass of water / Molar mass of water
= 0.029 kg / 18.02 g/mol
= 1.61 mol
Now, we need to determine the stoichiometric ratio between ethylene and ethanol in the reaction. The balanced chemical equation for the reaction is:
C2H4 + H2O → C2H5OH
From the balanced equation, we can see that one mole of ethylene reacts with one mole of water to produce one mole of ethanol. Therefore, the stoichiometric ratio between ethylene and ethanol is 1:1.
Since the stoichiometric ratio is 1:1, the limiting reagent is the reactant that is present in lesser moles. In this case, the limiting reagent is water (steam) because it is present in fewer moles than ethylene.
The maximum amount of ethanol that can be produced will be determined by the moles of water present. To calculate the maximum amount of ethanol, we will assume that all the water is consumed in the reaction.
3. Calculate the maximum moles of ethanol that can be produced:
- The moles of ethanol = Moles of water
= 1.61 mol
Finally, to calculate the maximum amount of ethanol in grams, we will multiply the moles of ethanol by its molar mass.
4. Calculate the maximum amount of ethanol:
- The molar mass of ethanol (C2H5OH) = 2 * atomic mass of carbon (C) + 6 * atomic mass of hydrogen (H) + atomic mass of oxygen (O)
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + 16.00 g/mol
= 46.07 g/mol
- Mass of ethanol = Moles of ethanol * Molar mass of ethanol
= 1.61 mol * 46.07 g/mol
= 74.29 g
Therefore, the maximum amount of ethanol that can be produced when 0.30 kg of ethylene and 0.029 kg of steam are placed into the reaction vessel is approximately 74.29 grams.