Diborane (B2H6), a useful reactant in organic synthesis, may be prepared by the following reaction?

Question

which occurs in a nonaqueous solvent: NaBH4 + BF3 -> B2H6 + NaBF4 (unbalanced). If the reaction has a 71% yield of diborane, how many grams of NaBH4 are needed to make 31.2 g B2H6?

Answer 1

The answer was:

4 BF3 + 3 NaBH4 → 2 B2H6 + 3 NaBF4 is the balanced equation.

3 moles of NaBH4 are needed for every 2 moles of B2H6.

31.2 g / 27.67 g/mol = 1.13 moles of B2H6 formed. If this is only 71 percent of the possible, then the possible moles was:

1.13 moles / 0.71 = 1.59 moles of B2H6

This would require 1.59 (3/2) = 2.39 moles of NaBH4.

2.39 moles (37.83 g/mol) = 90.41 grams of NaBH4.