how do you balance this type of molecular equation???

Question

Al(NO3)3(aq) + NaOH(aq)

i know it starts by putting the + and - together so you get AlOH+Na(NO3)3 for the second part. but does that mean i replace the original one? > NaOH???

after that how do you balance it and write the ionic equation??

i dont understand this at all. sorry.

Answer 1

First find the charges of the indiviual ions present:
Al = +3 NO3= -1 Na = +1 OH= -1

Keep the reactants the same, but Flip-Flop the negative ions in the products.

Al(NO3)3(aq) + NaOH (aq) -> Al(OH)3 (s) + NaNO3 (aq)

* note that it's Al(OH)3 because you're combining Al+3 and OH-1 and NaNO3 because it's Na+1 OH-1
** To find the phases of the 2 products use a solubility chart -- OH- are generally insoluble (except Group IA, *Ca2+, Ba2+, Sr2+) so that becomes a solid(your precipitate), NO3- are generally soluble so that will be aqueous.

Now you need to balance the equation

Al(NO3)3 + 3 NaOH --> Al(OH)3 + 3 NaNO3

*By adding the coefficients you now have the same number of each element on both sides of the arrow. Coefficients are multiplied by each element. Subscripts multiply the number of the element directly in front of them or all of the elements within paraentheses right before them: Reactants: Al - 3, N-3, 0-9, Na - 3, O-3, H - 3; products Al - 3, O - 3, H -3, Na - 3, N- 3, O- 9

To write the ionic equation take the equation you just wrote and break it apart into ions, be sure to keep coefficients, and subscripts become coefficients:

Al(NO3)3 becomes Al+3 and 3 NO3-1
3NaOH becomes 3Na+1 and 3 OH -1
Al(OH)3 is a solid, and therefore not broken apart into ions in solution
3 NaNO3 becomes 3Na+ and 3 NO3-1

Putting all that together:
Al+3(aq) + 3 NO3- (aq) + 3Na+1 (aq) + 3 OH -1(aq) --> Al(OH)3 (s) + 3Na+(aq) + 3 NO3-1(aq)

If you're also looking for the net ionic equation, you would cross out anything that appears EXACTLY the same on both sides of the equation. Here 3 NO3- (aq) and 3Na+1 (aq) Leaving:


Al+3(aq) + 3 OH -1(aq) --> Al(OH)3 (s)

(Hint: you'll end up with your solid and the 2 ions that made it up)

Answer 2

the balanced equation is: Al(NO3)3(aq)+ 3 NaOH(aq)= 3 NaNO3 +AL(OH)3. You were right about putting the + and - together to form the products however since Al is +3 and OH is -1 you need 3 OH to balance it out. Then you make a 'shopping list' and make sure whatever you have on one side equals what you have on the other side for all elements (Al, N, O, and Na). For this you need a 3 on NaOH and a 3 on NaNO3.

Answer 3

Al(NO3)3(aq)+NaOH(aq)= NaNO3+Al(OH)3


Al(NO3)3(aq)+ 3 NaOH(aq)= 3 NaNO3 +AL(OH)3
you have to watch out for indivual ions equivalency while wrighting the equation Al=3
then balance the equation normally.

Answer 4

Al(NO3)3+3NaOH==>Al(OH)3+3Na(NO3)

no u dont replace the original one

u write the oxidation state and balance the charges

Answer 5

3NaOH + Al(NO3)3 => 3NaNO3 + Al(OH)3 (s)
ionic equation
3OH- + Al3+ ==> Al(OH)3

Answer 6

Al(NO3)3(aq) + 3NaOH(aq) --> Al(OH)3(s) + 3NaNO3(aq)

Al3+(aq) + 3OH-(aq) --> Al(OH)3(s)

Answer 7

a. NH3(aq) + HNO3(aq) ---------> NH4NO3(aq) NH3(aq) + H3O+(aq) + NO3-(aq) --------> NH4+(aq) + NO3-(aq) + H2O(l) NH3(aq) + H3O+(aq) ---------> NH4+(aq) + H2O(l) b. Ba(OH)2(aq) + 2 HCl(aq) -------------> BaCl2(aq) + 2 H2O(l) Ba++(aq) + 2 OH-(aq) + 2 H3O+(aq) + 2Cl-(aq) ----> Ba++(aq) + 2Cl-(aq) + 4 H2O(l) OH-(aq) + H3O+(aq) ------------> 2 H2O(l) c. 3 HClO4(aq) + Fe(OH)3(s) -----------> Fe(ClO4)3 + 3 H2O(l) 3 H3O+(aq) + 3 ClO4-(aq) + Fe(OH)3(s) -----> Fe^+3(aq) + ClO4-(aq) + 6 H2O(l) 3 H3O+(aq) + Fe(OH)3(s) -----> Fe^+3(aq) + 6 H2O(l) d. AgOH(s) + HBr(aq) ------------> AgOH(s) + H2O(l) AgOH(s) + H3O+(aq) + Br-(aq) -----------> AgOH(s) + 2 H2O(l) AgOH(s) + H3O+(aq) + Br-(aq) -----------> AgOH(s) + 2 H2O(l)