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%

are there any other formulas for % yield other than actual yield/theoretical yield?

Answer 1

@UnkleRhaukus how do i find the limiting reagent?

Answer 2

d) 89 % is correct according to

http://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CCMQFjAA&url=http%3A%2F%2Fhome.comcast.net%2F~asewak%2FAPChemistry%2F3.rtf&ei=kTI6UMjnJMnYrQeh4oGADw&usg=AFQjCNHuXtEvWId0ydrtPTxh-qVj9ZfhtA&sig2=Br6X6jWMMRlXOjmqs839EA

Answer 3

lol nevermind...found it

Answer 4

@mathslover it isnt

Answer 5

I don't know whether the link is right or not

Answer 6

you're not yahoo. stop insisting

Answer 7

btw...you NEVER answered my question dont you take into consideration other factors other than just moles in deciding limiting reagent?

Answer 8

the limiting reagent is A

Answer 9

r u saying that to me @lgbasallote ?

Answer 10

to yahoo

Answer 11

that was actually my main question

Answer 12

oh I helped u a lot .. hence ok

Answer 13

and as usual...people tend to go into unnecessary details

Answer 14

@vishweshshrimali5 is excellent in this topic but he is not online .. :(

Answer 15

so anyway...tell me how to solve this

Answer 16

i get how the 4.5 was got

Answer 17

but why 4/4.5

Answer 18

lol

Answer 19

so we have

3 mol A

from the stoichiometry
2 mol A -> 3 mol C

so we will theoretically make
3/2 *3 mol c= 9/2 mol C

Answer 20

yield was 4 mol C

Answer 21

yes i got 9/2 too @UnkleRhaukus but why 4/9/2?

Answer 22

\[\text{relative % yield}=\frac{\text {yield}}{\text{theoretical yield}}\times100\%\]

Answer 23

ohh yes...how could i forget

Answer 24

\[=\frac{4}{9/2}\times100\%=800\%/9\approx89\%\]

Answer 25

my answer key still says 89 is wrong though

Answer 26

what is suggested by the key?

Answer 27

does the answer key's answer make sense?