I need help finding the net and total ionic for this reation. i know the steps but i keep getting the wrong answer.
Al2(SO4)3 (aq) + 6 NaOH (aq) ----> 2 Al(OH)3 (s) +3 Na2SO4 (aq)
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Al2(SO4)3 (aq) + 6 NaOH (aq) ---->
2Al(OH)3 (s) +3 Na2SO4 (aq)
1. Separate into ions. Solids stay as solids, aqueous solution are separated. (Gases, if you had one, would be written as the molecule, also).
2Al+3(aq) + 3SO4^-2(aq) + 6 Na^+(aq) + 6 OH^-(aq) ==> 2Al(OH)3(s) + 6Na^+(aq) + 3SO4^-2(aq)
Now cancel the ions common to both sides. 3SO4^-2 cancel. 6Na^+ cancel. You are left with
2Al^+3(aq) + 6OH^-(aq) ==> 2Al(OH)3(s)
Check my work. This is meticulous.
thanks.. i had probmels with the subscripts and coefficents. i didn't know you had to muliply it
The reaction given is a double replacement reaction, also known as a precipitation reaction. In this type of reaction, the positive and negative ions from two compounds switch places, resulting in the formation of two new compounds.
To find the net ionic equation, you need to eliminate the spectator ions. Spectator ions are the ions that are present on both sides of the equation and do not participate in the actual chemical reaction. They appear unchanged before and after the reaction.
Here are the steps to find the net ionic equation:
1. Write the balanced equation:
Al2(SO4)3 (aq) + 6 NaOH (aq) ----> 2 Al(OH)3 (s) + 3 Na2SO4 (aq)
2. Write the complete ionic equation by separating all the aqueous compounds into their respective ions:
2 Al3+ (aq) + 3 SO4^2- (aq) + 6 Na+ (aq) + 6 OH- (aq) ----> 2 Al(OH)3 (s) + 3 Na+ (aq) + 3 SO4^2- (aq)
3. Identify the spectator ions. In this case, Na+ and SO4^2- are the spectator ions because they are present on both sides of the equation without undergoing any change.
4. Write the net ionic equation by eliminating the spectator ions:
2 Al3+ (aq) + 6 OH- (aq) ----> 2 Al(OH)3 (s)
Now you have the net ionic equation, which shows only the species that actually participate in the reaction.
To find the total ionic equation, you simply write out all the ions present in the complete ionic equation.
Total ionic equation:
2 Al3+ (aq) + 3 SO4^2- (aq) + 6 Na+ (aq) + 6 OH- (aq) ----> 2 Al(OH)3 (s) + 3 Na+ (aq) + 3 SO4^2- (aq)
I hope this explanation helps. Let me know if you have any further questions!