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How is aluminium 3+?
Original post by BlueDiamondzz
How is aluminium 3+?
Can somebody please help me?
Al2(SO4)3
(SO4)2- has 2- charge
if Al in aluminium sulfate has ON x
then 2x + 3(-2) = 0 giving x = +3
(SO4)2- has 2- charge
if Al in aluminium sulfate has ON x
then 2x + 3(-2) = 0 giving x = +3
Original post by golgiapparatus31
Al2(SO4)3
(SO4)2- has 2- charge
if Al in aluminium sulfate has ON x
then 2x + 3(-2) = 0 giving x = +3
(SO4)2- has 2- charge
if Al in aluminium sulfate has ON x
then 2x + 3(-2) = 0 giving x = +3
I don’t understand the 3rd and 4th line of what you wrote can you please explain to me further
Original post by BlueDiamondzz
I don’t understand the 3rd and 4th line of what you wrote can you please explain to me further
Original post by BlueDiamondzz
I changed the ions to a 2- charge but it is still not +3 but now +2
Al can't have oxidation number 0 because then the charges don't balance in Al2(SO4)3
Original post by golgiapparatus31
Al can't have oxidation number 0 because then the charges don't balance in Al2(SO4)3
How do you calculate the charger of sulphur (sorry I’m really dumb). And what’s ON x?
Original post by BlueDiamondzz
How do you calculate the charger of sulphur (sorry I’m really dumb). And what’s ON x?
If we’re turning sulphur and oxygen it into an ion and there’s the remaining 1 do we transfer it to the aluminium to make it Al3 instead if Al2?
Original post by BlueDiamondzz
How do you calculate the charger of sulphur (sorry I’m really dumb). And what’s ON x?
The sulfate ion SO4^2- has a charge of 2-
The sulfur in it has an oxidation number (which I called ON) +6
I called the ON of aluminium x because it's unknown, then I solved for it. Did your teacher not explain this topic??
Original post by BlueDiamondzz
If we’re turning sulphur and oxygen it into an ion and there’s the remaining 1 do we transfer it to the aluminium to make it Al3 instead if Al2?
Sorry I don't understand..
Original post by golgiapparatus31
The sulfate ion SO4^2- has a charge of 2-
The sulfur in it has an oxidation number (which I called ON) +6
I called the ON of aluminium x because it's unknown, then I solved for it. Did your teacher not explain this topic??
Sorry I don't understand..
The sulfur in it has an oxidation number (which I called ON) +6
I called the ON of aluminium x because it's unknown, then I solved for it. Did your teacher not explain this topic??
Sorry I don't understand..
I understand now that x is 2x because aluminium is Al2. But I don’t understand why you did 3 x (-2). Is the -2 from the oxygen? And the 3 from the 3 part of the bracket of (SO4)3? Do you mind explaining it in steps and tell me if we need to turn it into an ion? Or maybe present your working out in a visual way. Sorry for bothering you. I have learning difficulties so I know it’s tough to explain things to me
Our teacher told us we turn it into an ion or something like that. I don’t remember and I thought I knew this topic but it turns out I don’t. I currently have no contact with the teacher so I can’t ask for help 😕
Original post by BlueDiamondzz
I understand now that x is 2x because aluminium is Al2. But I don’t understand why you did 3 x (-2). Is the -2 from the oxygen? And the 3 from the 3 part of the bracket of (SO4)3? Do you mind explaining it in steps and tell me if we need to turn it into an ion? Or maybe present your working out in a visual way. Sorry for bothering you. I have learning difficulties so I know it’s tough to explain things to me
Ok, don't worry about it!
Al2(SO4)3
The way I'm doing it is considering the Al and SO4 units separately
I know that SO42- has 2- charge.
The oxidation number of aluminium (I call it x because I need to calculate it, it is unknown) is equal to its charge
The overall charge is zero.
If I add charges of 2 Al(x+) units and 3 SO4(2-) units I get 0.
so 2x + 3(-2) = 0
2x = 6
x = +3
so the oxidation number is +3
Original post by BlueDiamondzz
I would also like to know the oxidation number for sulphur? I thought we times by the bracket for Al2(SO4)3 so I thought we would times 4 by 3
Sulphur is +2? And thank you 😁 for explaining this I think I have understood this now.
To calculate oxidation number of sulfur in sulfate (SO4)2-:
use the fact that you know the oxidation number of oxygen as -2
then call the oxidation number of S x
x + 4(-2) = -2
so x = +6
use the fact that you know the oxidation number of oxygen as -2
then call the oxidation number of S x
x + 4(-2) = -2
so x = +6
Original post by golgiapparatus31
To calculate oxidation number of sulfur in sulfate (SO4)2-:
use the fact that you know the oxidation number of oxygen as -2
then call the oxidation number of S x
x + 4(-2) = -2
so x = +6
use the fact that you know the oxidation number of oxygen as -2
then call the oxidation number of S x
x + 4(-2) = -2
so x = +6
I’m feeling confused again (sorry). I know that Al2(SO4)3 and we turn (SO4)3 into (SO4)2- then what happens to the remaining 1 or does that disappear because it’s a charge now?
Original post by BlueDiamondzz
I’m feeling confused again (sorry). I know that Al2(SO4)3 and we turn (SO4)3 into (SO4)2- then what happens to the remaining 1 or does that disappear because it’s a charge now?
You can't turn (SO4)3 into (SO4)2-.. that's turning S3O12 into SO4(2-) which makes no sense because atoms are not balanced and charges are not balanced. Not sure what you mean by remaining 1. I don't really understand your working.
Al2(SO4)3 is a shorthand way of writing [Al3+]2[SO42-]3
Original post by BlueDiamondzz
I’m feeling confused again (sorry). I know that Al2(SO4)3 and we turn (SO4)3 into (SO4)2- then what happens to the remaining 1 or does that disappear because it’s a charge now?
Is this right?
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