Question

When 3.54 grams of phosphorus react with excess oxygen gas to produce diphosphorus pentoxide, the percent yield is 84.1%. What was the actual yield of the reaction?

4P + 5O2 yields 2P2O5


5.46 g
6.82 g
9.10 g
9.64 g

Answer 1

@Compassionate

Answer 2

@vera_ewing

Answer 3

@uri

Answer 4

so whats the answers? @sweetburger

Answer 5

well can you solve for me please?

Answer 6

why

Answer 7

Its against ToS.

Answer 8

an no its not alot of other people give me the answers? can you just be kind and solve it

Answer 9

why your lying

Answer 10

common bro im being timed right now can you just solve it and be a live savor

Answer 11

ik my bad but please solve this for me

Answer 12

@sweetburger

Answer 13

can anyone just solve this

Answer 14

\[Percentage Yield = \frac{ Actual Yield }{ Theoretical Yield } \times 100\]
or
\[Percentage Yield = \frac{ Actual Yield \times 100 }{ Theoretical Yield }\]

Rearranging, we can get:
\[Actual Yield \times 100 = (Percentage Yield)(Theoretical Yield)\]
which means that
\[Actual Yield = \frac{ (Percentage Yield)(Theoretical Yield) }{ 100 }\]

From this rearranged equation, let's consider what we know and what we don't know based on the information given to us for the reaction:
- The percentage yield is 84.1% (or simply 84.1 in this case).
- The theoretical yield is unknown.

However, based on the balanced chemical equation for the reaction:
\[4P + 5O _{2} \rightarrow 2P _{2}O _{5}\]
we can work out the number of moles, and thus the number of grams of P2O5 we would EXPECT to form (i.e. our theoretical yield), given that we have reacted 3.54 g of phosphorus. The phosphorus is the limiting reagent whilst the other reactant, oxygen gas, is in excess, meaning that the amount of phosphorus we start with, and its molar ratio with the product we are forming, we be the determining factor into how much P2O5 can possibly be synthesised.

Here's a tutorial @taramgrant0543664 put together on how to go about finding the theoretical yield for a product from the balanced equation. One you have found it (in grams remember!), simply plug it in alongside our percentage yield value into the rearranged equation above to find the actual yield. Hope that helped you out a bit! :)
http://openstudy.com/study#/updates/559c49b4e4b0f93dd7c278ad

Answer 15

cn you guys just give me the answer @Ciarán95 @Cuanchi

Answer 16

Unfortunately I'm not going to do it out completely for you @jammy987...hopefully I've given you a basic setup for how to tackle the question. The key part here is working out the theoretical yield. Also, if you're required to answer a question like this, I presume it's something you need to know how to do yourself eventually.

Have you tried to work out the theoretical yield for the reaction yet...have you any ideas how you would go about it based on the link I left in my last post?

Answer 17

solveeee itttt

Answer 18

ok igot 8.10 so