pondering-soul
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Imagehow do they get the ionic equation?


thanks for any help in advance, I appreciate it
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andrei00
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1. Write a balanced symbol equation with state symbols.
2. Split ions if in aqueous solution(aq) , so Al2O3+ 2Na^+2OH^- + 3H2O = 2Na^+ + 2Al^3+ (OH)4^-
3. Cancel out “Spectator Ions” e.g ions present on both sides so Na+ ions on both sides so you remove them from the final equation.

PM or reply on here if you’re still struggling.

P.S this link might be helpful
https://www.onlinemathlearning.com/ionic-equation.html
Last edited by andrei00; 4 weeks ago
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pondering-soul
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(Original post by andrei00)
1. Write a balanced symbol equation with state symbols.
2. Split ions if in aqueous solution(aq) , so Al2O3+ 2Na^+2OH^- + 3H2O = 2Na^+ + 2Al^3+ (OH)4^-
3. Cancel out “Spectator Ions” e.g ions present on both sides so Na+ ions on both sides so you remove them from the final equation.

PM or reply on here if you’re still struggling.

P.S this link might be helpful
https://www.onlinemathlearning.com/ionic-equation.html
why do we not cancel out OH- ions aswell?
If there are 2OH- ions on left side, on right side there is 2(Al^3+ (OH)4 )^-1 ,since the overall molecule has a -1 charge, and Al has 3+ oxidation state, that would seem to suggest that there needs to be -4 more oxidation state , which is achieved through the (OH)4 , since each OH ion has -1 state and there are 4 ions. So can't it be written as Al^3+ 4OH^-1 (this would mean the charge on the molecule is -1, which it is ), and then cancel those OH from there?
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andrei00
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(Original post by pondering-soul)
why do we not cancel out OH- ions aswell?
If there are 2OH- ions on left side, on right side there is 2(Al^3+ (OH)4 )^-1 ,since the overall molecule has a -1 charge, and Al has 3+ oxidation state, that would seem to suggest that there needs to be -4 more oxidation state , which is achieved through the (OH)4 , since each OH ion has -1 state and there are 4 ions. So can't it be written as Al^3+ 4OH^-1 (this would mean the charge on the molecule is -1, which it is ), and then cancel those OH from there?
I assumed that the answer given in the thread starter was correct.

I think I figured it out but I might be wrong. So on the reactant side there is a 2- charge from the 2OH- and on the product side there’s also a 2- charge due to the 2Al+3= 6+ charge and (2OH)4- = 8- charge and 6+-8= 2- so both sides should be balanced. Thus the answer being Al2O3+2OH^- + 3H2O = 2Al^3+ 2(OH)4^-.

The OH- isn’t cancelled out because if you take 2OH- from (OH)4 that will equal (OH)2 and if we put the number of moles of it 2(OH)2=4OH and the number of H and O on the reactant side won’t be the same as the number of H and O on the product side.

Also, we would remove two OH- from the brackets rather than the number of moles because of BIDMAS(the correct order to complete an equation)Brackets, Indices, Division, Multiplication, Addition, Subtraction.
Last edited by andrei00; 4 weeks ago
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golgiapparatus31
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For step 2 in post 2,

the ions should be split as follows:
Al2O3+ 2Na^+2OH^- + 3H2O = 2Na^+ + 2[Al(OH)4]^-

The [Al(OH4)]^- ion is called the hydroxoaluminate ion
Last edited by golgiapparatus31; 4 weeks ago
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