Question

Ethanol is produced industrially by the acid catalyzed reaction of ethylene with water. The balanced equation for this reaction is:

C₂H₄(g) + H₂O(g) → C₂H₅OH(l)
If water is present in excess, how many grams of ethanol can be produced from 5.75 g of ethylene?

A) 265 g
B) 0.205 g
C) 3.50 g
D) 9.44 g

Answer 1

Using stoichiometry, 9.44 grams of ethanol can be produced from 5.75 g of ethylene, corresponding to option D.

Step-by-step explanation:

The question asks how many grams of ethanol can be produced from 5.75 g of ethylene given the balanced equation:

C2H4(g) + H2O(g) → C2H5OH(l)

To solve this problem, we need to perform a stoichiometric calculation. First, calculate the moles of ethylene using its molar mass and then use the balanced equation to find the moles of ethanol produced. Finally, convert moles of ethanol to grams using the molar mass of ethanol:

1. Calculate the moles of ethylene:
   
   Molar mass of ethylene (C2H4) = 28.05 g/mol
   Moles of ethylene = mass / molar mass = 5.75 g / 28.05 g/mol = 0.205 mol
2. According to the balanced reaction, 1 mole of ethylene produces 1 mole of ethanol. So, 0.205 mol of ethylene would produce 0.205 mol of ethanol.
3. Calculate the mass of ethanol produced:
   
   Molar mass of ethanol (C2H5OH) = 46.07 g/mol
   Mass of ethanol = moles of ethanol * molar mass of ethanol = 0.205 mol * 46.07 g/mol = 9.44 g

Therefore, 9.44 grams of ethanol can be produced from 5.75 g of ethylene, which corresponds to option D.