Question

Ethanol, C2H5OH, is produced industrially from ethylene, C2H4, by the following sequence of reactions: 3C2H4 + 2H2SO4 → C2H5HSO4 + (C2H5)2SO4 C2H5HSO4 + (C2H5)2SO4 + 3H2O → 3C2H5OH + 2H2SO4 What volume of ethylene at STP is required to produce 1.000 metric ton (1000 kg) of ethanol if the overall yield of ethanol is 90.1%

Answer 1

5.4 * 10⁵ L

Step-by-step explanation:

The series of reactions is:

• 3C₂H₄ + 2H₂SO₄ → C₂H₅HSO₄ + (C₂H₅)₂SO₄

• C₂H₅HSO₄ + (C₂H₅)₂SO₄ + 3H₂O → 3C₂H₅OH + 2H₂SO₄

We can see that both ethylene (C₂H₄) and ethanol (C₂H₅OH) have a stoichiometric coefficient of 3. In other words, one mol of ethylene equals 1 mol of ethanol.

Now we calculate the moles of ethanol that we want to produce:

• 1000 kg = 1000 * 10³ g = 10⁶ g

10⁶ g ethanol ÷ 46 g/mol = 2.1739 * 10⁴ mol ethanol

Calculating the moles of ethylene required, keeping in mind the yield:

2.174 * 10⁴ mol ethanol *
`(1molEthylene)/(1molEthanol)` * 100/90.1 = 2.4128* 10⁴ mol ethylene

Now we use PV=nRT to calculate V, because ethylene is a gas at STP. Using STP means that we have T = 273 K, and P = 1 atm.

1 atm * V = 2.4128* 10⁴ * 0.082 atm·L·mol⁻¹·K⁻¹ * 273 K

V = 5.4 * 10⁵ L