Chem 11 - Chemical Reactions?

Question

Under certain conditions, reacting 227.4 g of LiAlH4 with an excess of BF3 yields 93.84g of B2H6

Calculate the theoretical yield of B2H6.



What is the ACTUAL yield of B2H6?



Calculate the percentage yield of B2H6.

Answer 1

Supposing the reaction to be:
3 LiAlH4 + 4 BF3 → 2 B2H6 + 3 AlF3 + 3 LiF

(227.4 g LiAlH4) / (37.9543 g LiAlH4/mol) x (2 mol H2H6 / 3 mol LiAlH4) x (27.6696 g B2H6/mol) =
110.52 g = 110.5 g B2H6 in theory

You're told the actual yield: 93.84 g of B2H6

(93.84 g) / (110.52 g) = 0.8491 = 84.91% yield B2H6