Question

Consider the following reaction: 2A + B -> 3C + D 3.0 mol A and 2.0 mol B react to form 4.0 mol C. What is the percent yield of this reaction?

A) 50%
B) 67%
C) 75%
D) 89%
E) 100%

What other formulas for % yield other than actual/theoretical yield?

Answer 1

So the question or problem ask to compute the percentage yield of the reaction 2A+B->3C+D, so base on my calculation and further formulation and study about the said reaction, the answer among the following choices is letter D. 89%. I hope this would help 

Answer 2

Answer : The correct option is, (D) 89 %

Solution : Given,

Moles of A = 3.0 mol

Moles of B = 2.0 mol

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`2A+B\rightarrow 3C+D`

From the balanced reaction we conclude that

As, 2 mole of A react with 1 mole of B

So, 3 moles of A react with
`(3)/(2)=1.5` moles of B

From this we conclude that, B is an excess reagent because the given moles are greater than the required moles and A is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of C.

From the reaction, we conclude that

As, 2 mole of A react to give 3 mole of C

So, 3 mole of A react to give
`(3)/(2)* 3=4.5` mole of C

Theoretical yield of C = 4.5 mol

Actual yield of C = 4 mol

Now we have to calculate the percent yield of of the reaction.

`\% \text{ yield of reaction}=\frac{\text{ Actual yield of C}}{\text{ Theoretical yield of C}}* 100`

`\% \text{ yield of reaction}=(4mol)/(4.5mol)* 100=89\%`

Therefore, the percent yield of the reaction is 89 %