calculate the volume of ethene measured at room temperature and pressure which reacts to produce 1.0dm^3 of ethanol

Question

steam is reacted with ethene according to this equation:

C2H4 + H2O --> C2H5OH

ethanol has a density of 0.8kg/dm^3.

(Ar: C, 12; H, 1; O, 16.)

(at room temperature and pressure 1 mole of any gas has a volume of 24dm^3.)

show your working:

Answer 1

C₂H₄ + H₂O --> C₂H₅OH ... balanced equation

1.0 dm³ = 1 liter (1 L)

Let us convert 1.0 dm³ ethanol to moles of ethanol

1.0 dm³ x 0.8 kg/dm³ = 0.8kg = 800 g ethanol

molar mass ethanol = 2x12 + 6x1 + 1x16 = 24 + 6 + 16 = 46 g/mol

800 g x 1 mol/46 g = 17.39 moles ethanol

To produce 17.39 mols ethanol, according to the balanced equation, we would need 17.39 mols ethene

Converting this to volume, we have the following:

17.39 mols ethene x 24 dm³/mol = 417 dm³ = 400 dm³ (1 sig. fig.)