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%

5.4 * 10⁵ L

Explanation:

The series of reactions is:

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:

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 * * 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