Ethyl alcohol is prepared industrially by the reaction of ethylene, C2H4, with water. What is the percent yield of the reaction if 4.6 g of ethylene gives 4.7 g of ethyl alcohol?
C2H4(g) + H2O(l) = C2H6O(l)

% yield = 62.21 %

Explanation:

∴ mass C2H4(g) = 4.6 g

∴ mass C2H6O(l) = 4.7 g

theretical yield:

∴ molar mass C2H4(g) = 28.05 g/mol

⇒ mol C2H4(g) = (4.6 g)*(mol/28.05 g) = 0.164 mol

⇒ mol C2H6O(l) = (0.164 mol C2H4)*(mol C2H6O(l)/molC2H4(g))

⇒ mol C2H6O(l) = 0.164 mol

∴ molar mass C2H6O(l) = 46.07 g/mol

⇒ mass C2H6O(l) = (0.164 mol)*(46.07 g/mol) = 7.55 g

⇒ theoretical yield = 7.55 g

⇒ % yield = (4.7 g)/(7.55 g))*100

⇒ % yield = 62.21 %