VCAA 2011, SA, 4)a. Don't understand stoichiometric ratio

[bentennason]

The solutions state that n(P2O5)= n(MgNH4PO4.6H2O) / 2
Why is it half?

[rebeckab]

you need twice as much as the second one to get the amount of Ps you need for the first one, so the first is half as many as the second one

[vea]

^
Sometimes you just gotta look at things in the most basic way instead of trying to figure out the equation for it.

This one got me good last year but don't fret, examiners give you consequentials for the rest of the working out, they only -1'd me for this!

[anthony1n]

Since all the P205 is present in MgNH4PO4.6H2O, there needs to be a 2 infront of the MgNH4PO4.6H2O to balance the equation.
So : P205 --> 2MgNH4PO4.6H2O
Now you can use the mol ratio to find the mole of P2O5
n(P2O5)/n(MgNH4PO4.6H2O) = 1/2
n(P2O5)= n(MgNH4PO4.6H2O) / 2

[bentennason]

ah okay, thanks brewww

[ggxoxo]

Can someone explain part b of this qn; how can we assume that by dividing by a lower molar mass we get the same amount in mol?

[Mao]

Can someone explain part b of this qn; how can we assume that by dividing by a lower molar mass we get the same amount in mol?

Let me use a simpler example to illustrate this point:

Suppose we have an impure sample of A. We then go through a whole crapload of reaction to get , we weight this, and get some number of moles.

Now, upon heating, , with the excess X fucking off somewhere¹. We have a total conversion, we have a stoichiometric conversion. Since this is a 1-to-1 reaction, our number of moles have not changed.

As the molar mass of the product has decreased, so too will the mass. But their ratio mass/molar-mass is the same, because that's the number of moles, and we still have the same number of moles.

^^same principles.


¹ Yeah, that is a scientific term, bitches.

[ligands]

does that mean you can just go p2o5 -> 2po4?

[Mao]

does that mean you can just go p2o5 -> 2po4?

yes. It implies: , which is perfectly fine for the calculation we're trying to do here (gravimetric analysis).