Diborane is an important reagent for Organic Chemistry Synthesis. It may be prepared by the following reaction:

Question

anonymous · Answer

$$2NABH_4(s)+4BF_3(g) \rightarrow 2B_2H_6(g)+3NaBF_4(s)$$
If the reaction has a 75% yield of Diborane, how many grams of NaBH_4 are needed to make 20.0g of $$B_2H_6$$?

anonymous · Answer

75% of the theoretical yield of this reactions is 20 grams. So, in theory the reaction is supposed to produce 26.7 grams of B2H6. In other words, if the reaction would proceed at 100% yield, we would obtain 26.7 grams but in practice we only get 20 grams. 

Then, we need to find out what amount of reagents are necessary to produce 27 grams of diborane. We will have to calculate the moles. 

The molecular mass of B2H6 is 26.7. This means that we have $$n=\frac{ m }{ M }=\frac{ 26.7 }{ 26.7 } = 1mol $$
The chemical equation says that n(NaBH4) : n(B2H6) = 2 : 2 = 1 : 1

As n(B2H6) = 1 mol, n(NaBH4) = 1 mol. The molecular mass of the latter is 38, so

$$m(NaBH _{4})=nM=1.38 = 38 g$$

anonymous · Answer

Did I explain it clearly?

anonymous · Answer

Yes you did. I'll make some corrections because I made a mistake in the balancing. I meant to put "3" rather than "2" in front of NaBH4. Other than that, it looks good! Thanks, @suzanka!!

anonymous · Answer

Oops, I didn't notice the equation wasn't balanced correctly. Good that you noticed :)