anonymous
  • anonymous
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
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@Rushwr
anonymous
  • anonymous
@Data_LG2
owlet
  • 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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anonymous
  • anonymous
@owlet could you show me through a equation on what to do?
anonymous
  • anonymous
@ganeshie8
owlet
  • 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
  • anonymous
Okay so 8.07 g * 219.8 * (4/5) ?
anonymous
  • anonymous
Is that right?
anonymous
  • anonymous
@Photon336
Photon336
  • Photon336
let's balance the reaction first
anonymous
  • anonymous
It's already balanced isn't it?
anonymous
  • anonymous
5P4O6 + 8I2 --> 4P2I4 + 3P4 O10
Photon336
  • Photon336
\[5P_{4}O_{6} + 8I_{2} --> 4P_{2}I_{4} + 3P_{4}O_{10}\]
Photon336
  • Photon336
Good
Photon336
  • Photon336
Hey can you find the molar masses of all these compounds first before we even do anything.
anonymous
  • anonymous
P4O6 = 219.8 I2 =126.9
Photon336
  • Photon336
@staldk3 the first thing we must do is convert both to moles. |dw:1446876925755:dw|
Photon336
  • Photon336
This is what we get 0.037 moles of P4O6 0.079 moles of I2
Photon336
  • 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
  • anonymous
So I would take .037mol * (8/5) = .079 * (5/8) =
anonymous
  • anonymous
Is that right?
Photon336
  • Photon336
@staldk3
Photon336
  • Photon336
I want you to get into the habit of doing this
Photon336
  • 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
  • Photon336
Can you show me how much P4O6 we need based on what I did?
Photon336
  • Photon336
well use the same method I did and show me
anonymous
  • anonymous
|dw:1446877876281:dw|
Photon336
  • Photon336
If you do it like that you'll never get confused
Photon336
  • Photon336
Now let's make a chart
anonymous
  • anonymous
Um for the P4O6 I tried it out and got .0592
Photon336
  • Photon336
that should be .0592 for the first one |dw:1446878011348:dw|
Photon336
  • Photon336
@staldk3 look at what I did for the chart
Photon336
  • 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
  • anonymous
@Photon336 So I2 is the limiting reactant?
Photon336
  • Photon336
how do you know this @staldk3
anonymous
  • anonymous
Because it has less.
Photon336
  • Photon336
We HAVE 0.037 moles but we NEED 0.0594 MOLES
anonymous
  • anonymous
We only have .037 of I2 but need .0592
Photon336
  • 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
  • anonymous
We use the I2 amount to figure out how much product we can make.
Photon336
  • Photon336
yes. ill show you how
anonymous
  • anonymous
Thank you
Photon336
  • Photon336
|dw:1446879046044:dw|
Photon336
  • Photon336
why must we multiply by this molar ratio and why is I2 in the denominator @staldk3 ?
anonymous
  • anonymous
to cancel out
Photon336
  • Photon336
this is how you must set up your ratios
Photon336
  • Photon336
so can you show me how to get the number of moles of the product we want?
Photon336
  • 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
  • anonymous
Okay so I take I2 .037 I2 and multiple it (4P2I4 / 8 I2)
anonymous
  • anonymous
|dw:1446879436572:dw|
Photon336
  • 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
  • Photon336
now from this you can easily calculate the number of grams produced
anonymous
  • anonymous
.0185 P2I4
Photon336
  • Photon336
Most importantly though do you get the pattern?
anonymous
  • anonymous
Sorta
Photon336
  • 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
  • anonymous
I would use .037 for this value as well right?
Photon336
  • Photon336
0.0185 is the number of moles you found for P2I4
Photon336
  • Photon336
So you need to use that and multiply it by its molar mass.
Photon336
  • Photon336
does that make sense?
anonymous
  • anonymous
Where did the 0.0185 come from?
Photon336
  • Photon336
|dw:1446880054387:dw|
anonymous
  • anonymous
Oh okay.
anonymous
  • anonymous
|dw:1446880135102:dw|
anonymous
  • anonymous
@Photon336
Photon336
  • Photon336
Yeah that's how you get 0.0185 moles of P2I4
Photon336
  • Photon336
Then you multiply that by the molar mass of P2I4
anonymous
  • anonymous
The .0185 or the .037?
Photon336
  • Photon336
|dw:1446880578551:dw|
Photon336
  • Photon336
that's all you need to do then to find the mass you would get for this compound
anonymous
  • anonymous
Okay so 219.8 g P2I4 * .0185
Photon336
  • Photon336
\[0.0185 mol*(\frac{ 570 grams}{ mol P_{2}I_{4}} )= 10.6 grams P_{2}I_{4}\]
anonymous
  • anonymous
Where is the 570 grams coming from?
Photon336
  • Photon336
That's the molar mass of P2I4 570grams/mole. remember? it's a different compound
anonymous
  • anonymous
Alright. Gotcha.
Photon336
  • 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
  • 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
  • Photon336
I think you should always make a chart keeping track of what is what
anonymous
  • anonymous
The 10.6 is the max?
Photon336
  • Photon336
yes because remember that's based off of the stoichiometry
Photon336
  • Photon336
But sometimes in lab you dont always get the amount that you should get
Photon336
  • Photon336
that's why they call the amount/ mass you get from the balanced equation the theoretical yield
anonymous
  • anonymous
It's counting the 10.6 wrong for some reason.
Photon336
  • 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
  • Photon336
Oh crap. wait I see
anonymous
  • anonymous
|dw:1446881527863:dw|
Photon336
  • 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
  • anonymous
Okay so the max is actually 16.9?
Photon336
  • Photon336
Yeah unless your question requires significant figures.
anonymous
  • anonymous
I would say it does.
anonymous
  • anonymous
It's 3 sig figs right?
Photon336
  • Photon336
I think we had 3 so we need to end off with three
Photon336
  • Photon336
that i'm not so sure about
Photon336
  • Photon336
I g2g let me know how it goes
anonymous
  • anonymous
Actually I think it would be 17 since 570 is the smallest and I don't think the zero counts.
anonymous
  • anonymous
Thanks for the help!
Photon336
  • Photon336
No problem

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