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 1.0 kg of ethylene (C2H4) and 0.014 kg of steam are placed into the reaction vessel?

This is a limiting reagent problem. How do I know that? Because amounts of BOTH reactants are given. I solve these problems by solving two stoichiometry problems. First I use 1000g ethylene and all of the steam needed and calculate moles ethanol produced. Then I use 14 g steam and all of the ethylene needed and calculate the moles ethanol produced. There is a different answer produced and both answers can't be correct; however, in limiting reagent problems the smaller value is ALWAYS the correct one and the reagent producing that value is the limiting reagent. Here is a worked example of a stoichiometry problem. Just follow the steps. Post your work if you get stuck.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To find the maximum amount of ethanol that can be produced, we need to determine which reactant limits the amount of product formed. This can be done by calculating the number of moles of each reactant and comparing their stoichiometric ratios.

First, let's convert the masses of ethylene (C2H4) and steam (H2O) into moles using their respective molar masses:

Molar mass of C2H4:
C = 12.01 g/mol
H = 1.01 g/mol
Molar mass of C2H4 = (2 * C) + (4 * H) = 2 * 12.01 + 4 * 1.01 = 28.05 g/mol

Moles of C2H4 = Mass of C2H4 / Molar mass of C2H4
Moles of C2H4 = 1000 g / 28.05 g/mol ≈ 35.69 mol

Molar mass of H2O:
H = 1.01 g/mol
O = 16.00 g/mol
Molar mass of H2O = (2 * H) + O = 2 * 1.01 + 16.00 = 18.02 g/mol

Moles of H2O = Mass of H2O / Molar mass of H2O
Moles of H2O = 14 g / 18.02 g/mol ≈ 0.777 mol

Now let's look at the balanced equation for the reaction:

C2H4(g) + H2O(g) → C2H5OH(l)

From the balanced equation, we can see that the stoichiometric ratio between C2H4 and C2H5OH is 1:1. This means that 1 mole of C2H4 can produce 1 mole of C2H5OH.

Since we have 35.69 moles of C2H4, we can produce a maximum of 35.69 moles of C2H5OH.

Now, let's calculate the mass of ethanol (C2H5OH) produced using its molar mass:

Molar mass of C2H5OH:
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Molar mass of C2H5OH = (2 * C) + (6 * H) + O = (2 * 12.01) + (6 * 1.01) + 16.00 = 46.07 g/mol

Mass of C2H5OH = Moles of C2H5OH * Molar mass of C2H5OH
Mass of C2H5OH = 35.69 mol * 46.07 g/mol ≈ 1643.38 g

Therefore, the maximum amount of ethanol (C2H5OH) that can be produced is approximately 1643.38 grams.