What is the maximum mass of p2i4 that can be prepared from 8.07 g of p4o6 and 9.97 g of iodine according to the reaction 5p4o6 + 8i2 yields 4p2i4 + 3p4o10

- anonymous

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- schrodinger

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- anonymous

@Rushwr

- anonymous

@Data_LG2

- owlet

I think you have to convert the given masses of the reactants into moles first. Then using mole ratios, you can be able to determine which will yield the least amount of product, which will be your limiting reagent. The limiting reagent will determine how much of the product will be produced.

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## More answers

- anonymous

@owlet could you show me through a equation on what to do?

- anonymous

@ganeshie8

- owlet

Step 1:
given mass of p4o6 x molar mass of p4o6 x (4 mole p2i4/ 5 p4o6 ) = ____ moles of p2i4
given mass of i2 x molar mass of i2 x ( 4 mole p2i4/ 8 mole i2 ) = _____ moles of p2i4
Step 2:
use the smallest amt of moles of p2i4 from step 1 to solve for the mass of p2i4
_____ moles of p2i4 x (molar mass of p2i4) = _____g of p2i4 --> this is the value that the question is asking.

- anonymous

Okay so 8.07 g * 219.8 * (4/5) ?

- anonymous

Is that right?

- anonymous

@Photon336

- Photon336

let's balance the reaction first

- anonymous

It's already balanced isn't it?

- anonymous

5P4O6 + 8I2 --> 4P2I4 + 3P4 O10

- Photon336

\[5P_{4}O_{6} + 8I_{2} --> 4P_{2}I_{4} + 3P_{4}O_{10}\]

- Photon336

Good

- Photon336

Hey can you find the molar masses of all these compounds first before we even do anything.

- anonymous

P4O6 = 219.8
I2 =126.9

- Photon336

@staldk3 the first thing we must do is convert both to moles.
|dw:1446876925755:dw|

- Photon336

This is what we get
0.037 moles of P4O6
0.079 moles of I2

- Photon336

To find the limiting reagent we must multiply both the number of moles we have for each reactant by the molar ratio to find out how much we need for the reaction.
Make sense?

- anonymous

So I would take .037mol * (8/5) =
.079 * (5/8) =

- anonymous

Is that right?

- Photon336

@staldk3

- Photon336

I want you to get into the habit of doing this

- Photon336

When you multiply by the molar ratio always include the compounds in there too like as I did below. if you wanted to find out say how much I2 you need you would write the ratio such that P4O6 is in the denominator
|dw:1446877619451:dw|

- Photon336

Can you show me how much P4O6 we need based on what I did?

- Photon336

well use the same method I did and show me

- anonymous

|dw:1446877876281:dw|

- Photon336

If you do it like that you'll never get confused

- Photon336

Now let's make a chart

- anonymous

Um for the P4O6 I tried it out and got .0592

- Photon336

that should be .0592 for the first one
|dw:1446878011348:dw|

- Photon336

@staldk3 look at what I did for the chart

- Photon336

once you identify the limiting reagent then you can multiply by the molar ratio to find how much product you need or the moles of that product.

- anonymous

@Photon336 So I2 is the limiting reactant?

- Photon336

how do you know this @staldk3

- anonymous

Because it has less.

- Photon336

We HAVE 0.037 moles but we NEED 0.0594 MOLES

- anonymous

We only have .037 of I2 but need .0592

- Photon336

so we automatically know that Iodine runs out FIRST. so NoW. what do you think we must do now to finish this problem?

- anonymous

We use the I2 amount to figure out how much product we can make.

- Photon336

yes. ill show you how

- anonymous

Thank you

- Photon336

|dw:1446879046044:dw|

- Photon336

why must we multiply by this molar ratio and why is I2 in the denominator @staldk3 ?

- anonymous

to cancel out

- Photon336

this is how you must set up your ratios

- Photon336

so can you show me how to get the number of moles of the product we want?

- Photon336

be sure to do it the same way I wrote above, so that you can see for yourself what compounds will cancel out.

- anonymous

Okay so I take I2 .037 I2 and multiple it (4P2I4 / 8 I2)

- anonymous

|dw:1446879436572:dw|

- Photon336

Yes exactly do the most important pattern I want you to see is that Iodine cancels out because it's in the denominator.
For the units
\[I_{2}*\frac{ p_{2}I_{4} }{ I_{2} }\]

- Photon336

now from this you can easily calculate the number of grams produced

- anonymous

.0185 P2I4

- Photon336

Most importantly though do you get the pattern?

- anonymous

Sorta

- Photon336

so for this all you need to do now is multiply by the molar mass of P2I4 to get the number of grams of P2I4

- anonymous

I would use .037 for this value as well right?

- Photon336

0.0185 is the number of moles you found for P2I4

- Photon336

So you need to use that and multiply it by its molar mass.

- Photon336

does that make sense?

- anonymous

Where did the 0.0185 come from?

- Photon336

|dw:1446880054387:dw|

- anonymous

Oh okay.

- anonymous

|dw:1446880135102:dw|

- anonymous

@Photon336

- Photon336

Yeah that's how you get 0.0185 moles of P2I4

- Photon336

Then you multiply that by the molar mass of P2I4

- anonymous

The .0185 or the .037?

- Photon336

|dw:1446880578551:dw|

- Photon336

that's all you need to do then to find the mass you would get for this compound

- anonymous

Okay so 219.8 g P2I4 * .0185

- Photon336

\[0.0185 mol*(\frac{ 570 grams}{ mol P_{2}I_{4}} )= 10.6 grams P_{2}I_{4}\]

- anonymous

Where is the 570 grams coming from?

- Photon336

That's the molar mass of P2I4 570grams/mole. remember? it's a different compound

- anonymous

Alright. Gotcha.

- Photon336

This is the maximum mass we would get for that compound provided that there aren't any side reactions or mishaps in the lab.

- Photon336

If you want to practice I suggest trying this one on your own
Say we have this reaction
3A + B --> C
6 grams of A
5 grams of B
Molar mass of B = 2grams/mol
Molar mass of A = 1gram/mol
Molar mass of C = 4 grams/mol.

- Photon336

I think you should always make a chart keeping track of what is what

- anonymous

The 10.6 is the max?

- Photon336

yes because remember that's based off of the stoichiometry

- Photon336

But sometimes in lab you dont always get the amount that you should get

- Photon336

that's why they call the amount/ mass you get from the balanced equation the theoretical yield

- anonymous

It's counting the 10.6 wrong for some reason.

- Photon336

the value you get in the lab is called the experimental yield. the value you get like the 10.6 is the theoretical yield. the percent yield is experimental yield / theoretical x 100% and basically the closer experimental yield is to the theoretical yield the better your experiment is.
|dw:1446881289752:dw|

- Photon336

Oh crap. wait I see

- anonymous

|dw:1446881527863:dw|

- Photon336

I made a minor error earlier. I believe this is the correct answer.
We have this
0.037 moles of P4O6
0.079 moles of I2
Moles needed for reach
0.049 mol P4O6 required
0.054 mol I2 required
the limiting reagent was P4O6
\[0.037 mol*\frac{ 4P_{2}I_{4} }{ 5P_{4}O_{6} } = 0.0296*570g = 16.87 grams\]

- anonymous

Okay so the max is actually 16.9?

- Photon336

Yeah unless your question requires significant figures.

- anonymous

I would say it does.

- anonymous

It's 3 sig figs right?

- Photon336

I think we had 3 so we need to end off with three

- Photon336

that i'm not so sure about

- Photon336

I g2g let me know how it goes

- anonymous

Actually I think it would be 17 since 570 is the smallest and I don't think the zero counts.

- anonymous

Thanks for the help!

- Photon336

No problem

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