VCAA 2011 Q4a) need help

[soccerboi]

http://www.vcaa.vic.edu.au/vcaa/vce/studies/chemistry/exams/2011/2011chem1-w.pdf

I can't get my head around why n(P2O5) is half the mol of (MgNH4PO4.6H2O). i thought it would be twice the mol, since its P2 compared to just P in the precipitate.

[joseph95]

You would write it out as P2O5 = 2MgNH4PO4.6H2O to balance the Ps.
Using mol ratio, n(P2O5) = (1/2) * n(MgNH4PO4.6H2O)

[soccerboi]

How would you write the chemical equation between P2O5, NH3 and MgSO4.7H2O?
I think i wrote the equation without correctly balancing it ...

[Aurelian]

If I have 10 single oranges, and you repackage them into sets of two oranges, you will have 5 sets of two oranges.

If I have 10 mol of MgNH4PO4.6H2O, but it is present as P2O5 originally, I would have had 5 mol of P2O5. I am repackaging 'single' bits of phosphorus into packets of two. Does that help?

[joseph95]

When I first saw that question I was so confused as to how to write that too. But the only information you need is n(P2O5) = (1/2) * n(MgNH4PO4.6H2O). Everything else is a distractor - read the assessor's report.

"Many students struggled to correctly identify the data relevant to the calculations and attempted to use all data."

[charmanderp]

IIRC it says somewhere in the question that all the phosphorous present in the precipitate is present in (MgNH4PO4.6H2O) and all the phosphorous present in the original sample is present in P2O5. So therefore the amount of phosphorous present in either is the same.

So to calculate the amount of P2O5 you take the ratio between phosphorous and P2O5, which is 2:1, and then divide accordingly.

You don't really even need to write an equation, just understand the written question.

[ligands]

You don't need to balence the entire equation if there is only 2 sets of p, you would just separate the equations to equal p2o5 -> po4 and balance

[aznxD]

n(P2O5) = 1/2*n(P) = 1/2*n(MgNH4PO4.6H2O)
For example, if you were given 1 mol of P. You can only form 1/2 mol of P2O5 as each P2O5 molecule has 2 P atoms.

[soccerboi]

You don't need to balence the entire equation if there is only 2 sets of p, you would just separate the equations to equal p2o5 -> po4 and balance

So whenever we're given something complicated, we can just grab the bit we want, in this case the compounds concerning P, and just balance that bit, without balancing the whole eqn and still get the correct mol ratios?

[ligands]

You don't need to balence the entire equation if there is only 2 sets of p, you would just separate the equations to equal p2o5 -> po4 and balance

So whenever we're given something complicated, we can just grab the bit we want, in this case the compounds concerning P, and just balance that bit, without balancing the whole eqn and still get the correct mol ratios?

Yes only if it is something like this through with the p2o5 out you can then just grab the p's in the long equation, only sometimes in grav do you need to balence the entire equation

[jaydee]

the easiest way to look at it is n(MgNH4PO4.6H2O) = n(P) since theres only one phosphurus in the compound. this comes the phosphurus in P2O5. P2O5 --> 2P + 5O. the n(P) is what u know so n(P2O5) = 0.5 x n(P). Obviously P2O5 isnt a reaction but it's the same concept for all graviemtric analyses when ur converting ions from one form to another

[Tonychet2]

quite simply, because to balance the P atoms there are 2 in the first compound (P2O5) then there in the next compound there is ONE

so the mole ratio  n(P2O5) = 2*n(MgNH4PO4.6H2O)

sorry to clear the ambiguity of this i meant n(P2O5) = 2n(MgNH4PO4.6H2O), didnt realise putting a * would change the meaning sry

edit: again oops ignore this because it was interpreted as n(p2o5) = twice the amount of n(MgNH4PO4.6H2O)

[charmanderp]

quite simply, because to balance the P atoms there are 2 in the first compound (P2O5) then there in the next compound there is ONE

so n(P2O5) = 2*n(MgNH4PO4.6H2O)

The n(P2O5) is actually half the n(MgNH4PO4.6H2O), rather than double. Because n(MgNH4PO4.6H2O) is equal to n(P) in MgNH4PO4.6H2O, and n(P2) in P2O5 is equal to n(MgNH4PO4.6H2O).

[Tonychet2]

quite simply, because to balance the P atoms there are 2 in the first compound (P2O5) then there in the next compound there is ONE

so n(P2O5) = 2*n(MgNH4PO4.6H2O)

The n(P2O5) is actually half the n(MgNH4PO4.6H2O), rather than double. Because n(MgNH4PO4.6H2O) is equal to n(P) in MgNH4PO4.6H2O, and n(P2) in P2O5 is equal to n(MgNH4PO4.6H2O).

yes i know , but instead of writing 1/2n(P2O5) = n(MgNH4PO4.6H2O)
i multiplied each side by 2 so it would be easier to read / make more sense either way its the same

1/2n(P2O5) = n(MgNH4PO4.6H2O) is the same as  n(P2O5) = 2*n(MgNH4PO4.6H2O)

edit: just realised i made myself ambiguous by putting the * between the 2 and n(MgNH4PO4.6H2O) i meant to write n(P2O5) = 2n(MgNH4PO4.6H2O)

[Tonychet2]

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