Question

21 g of butylene reacts with 115 ml of water and a sulfuric catalyst at 77 C to produce 10.7g of butanol. Calculate the percent yield for the reaction. Reminder: the density of water changes with increasing temperature.

Answer 1

Butylene + H2O --------H+-----------> Butanol

C4H8 + H2O --------H+------------> C4H10O

Mass of butylene = 21 g.

Molar mass of butylene = 56 g./mol

Amount of butylene = mass / molar mass = 21 / 56 = 0.375 mol

Mass of water = density * volume = 0.9820 * 115 = 112.93 g.

Molar mass of water = 18. g/mol

Amount of water = mass / molar mass = 112.93 / 18 = 6.27 mol

From the balanced equation, 1 mol butylene = 1 mol water

So, butylene is limiting reagent.

Again from the balanced equation,

1 mol butylene forms 1 mol butnol

then, 0.375 mol butylene forms 0.375 mol butanol

Theoretical mass of butanol = moles * molar mass = 0.375 * 64. = 24.0 g.

but actual mass of butanol = 10.7 g.

% yield = actual mass * 100 / theoretical yield = 10.7 * 100 / 24.0 = 44.6 %