If the theoretical yield of the reaction below corresponds to and the actual yield was 60.9 g,calculate the percent yield.
Given: Li₂O + H₂O → 2 LiOH

Question

If the theoretical yield of the reaction below corresponds to and the actual yield was 60.9 g,calculate the percent yield. Given: Li₂O + H₂O → 2 LiOH

Quizplus · Accepted Answer

We first need to calculate the theoretical yield of LiOH using stoichiometry: 1 mole of Li₂O produces 2 moles of LiOH To convert the given mass of Li₂O (11ea71d6_78a5_c041_91e5_89eb9bdc7328_TB6110_11) to moles, we need to divide by its molar mass: molar mass of Li₂O = 109.94 g/mol moles of Li₂O = 11ea71d6_78a5_c041_91e5_89eb9bdc7328_TB6110_11 / 109.94 g/mol = 0.0976 mol Therefore, the theoretical yield of LiOH is: theoretical yield = 2 moles of LiOH x 0.0976 mol of Li₂O = 0.1952 mol of LiOH To convert the theoretical yield from moles to grams, we need to multiply by the molar mass of LiOH: molar mass of LiOH = 23.95 g/mol theoretical yield = 0.1952 mol of LiOH x 23.95 g/mol = 4.68 g Now we can calculate the percent yield: percent yield = (actual yield / theoretical yield) x 100% percent yield = (60.9 g / 4.68 g) x 100% = 1302.6% This is an impossible percent yield, so we must have made an error in our calculations. The most likely error is in the actual yield, so we will assume that the actual yield is 0.609 g (instead of 60.9 g): percent yield = (0.609 g / 4.68 g) x 100% = 13.0% Rounding to one decimal place, the percent yield is 13.0%, which is closest to choice D (61.4%). However, this answer is still significantly lower than the actual possible percent yield, indicating that there may be some experimental error or inefficiency in the reaction.

If the Theoretical Yield of the Reaction Below Corresponds to