Chemistry- question?

Question

If 455 g Be(OH)2 were to react with 835 g V(CN)3 , and 437 g of Be(CN)2 and some V(OH)3 were formed, what would be the percent yield of this reaction?

Answer 1

Atomic weights: Be=9 O=16 H=1 C=12 N=14 Be(OH)2=43 Be(CN)2=61

3Be(OH)2 + 2V(CN)3 ===. 3Be(CN)2 + 2V(OH)3

Because of the wording of the problem, Be(OH)2 is the limiting reagent, and one doesn't worry about V(CN)3.

455gBe(OH)2 x 1molBe(OH)2/43gBe(OH)2 x 3molBe(CN)2)/3molBe(OH)2 x 61gBe(CN)2/1molBe(CN)2 = 645g Be(CN)2

437gBe(CN)2/645gBe(CN)2 x 100% = 67.8%