How many moles of NaBH4 are needed to generate one mole of B2H6 if the reaction gives an 80% yield?
3NaBH4 +4BF3 �¨ 3NaBF4 + 2B2H6
To determine how many moles of NaBH4 are needed to generate one mole of B2H6, we must first compare the stoichiometry of these two compounds in the balanced chemical equation.
The balanced chemical equation given is:
3NaBH4 + 4BF3 -> 3NaBF4 + 2B2H6
From the equation, we can see that 3 moles of NaBH4 react to form 2 moles of B2H6. This is evident from the coefficient ratio between NaBH4 and B2H6, which is 3:2.
Since the reaction has an 80% yield, we need to consider the actual amount of NaBH4 required to produce one mole of B2H6. If the yield were 100%, 3 moles of NaBH4 would be required. However, since the yield is only 80%, the actual amount needed is:
Actual Amount = Theoretical Amount / Yield
Actual Amount = (3 moles / 2 moles) / 0.80
Actual Amount = 1.875 moles of NaBH4
Therefore, approximately 1.875 moles of NaBH4 are needed to generate one mole of B2H6 with an 80% yield.