Asked by Jess
For each reaction, complete and balance the equation and give the net ionic equation.
a) KOH(aq)+ZnCl2(aq)-> KCl2+ZnOH
The second half of the equation is how i had done on the exam but he said its not balanced. How do I balance this? How would it look. I keep trying but not getting. I would appreciate it if you could do this question so I can use as a guide for other similar questions.
a) KOH(aq)+ZnCl2(aq)-> KCl2+ZnOH
The second half of the equation is how i had done on the exam but he said its not balanced. How do I balance this? How would it look. I keep trying but not getting. I would appreciate it if you could do this question so I can use as a guide for other similar questions.
Answers
Answered by
DrBob222
Your problem is you don't have the correct formula for zinc hydroxide. It should be Zn(OH)2.
2KOH(aq) + ZnCl2(aq) ==> 2KCl(aq) + Zn(OH)2(s)
You can put in the aq (and s) on the total ionic equation but here it is.
2K^+ + 2OH^- + Zn^+2 + 2Cl^- ==>2K^+ + 2Cl^- + Zn(OH)2
Now cancel the ions common to both sides; i.e., 2K^+ appears on both sides and 2Cl^- appears on both sides. Get rid of them and you are left with
2OH^-(aq) + Zn^+2(aq) ==> Zn(OH)2(s) which is the net ionic equation.
2KOH(aq) + ZnCl2(aq) ==> 2KCl(aq) + Zn(OH)2(s)
You can put in the aq (and s) on the total ionic equation but here it is.
2K^+ + 2OH^- + Zn^+2 + 2Cl^- ==>2K^+ + 2Cl^- + Zn(OH)2
Now cancel the ions common to both sides; i.e., 2K^+ appears on both sides and 2Cl^- appears on both sides. Get rid of them and you are left with
2OH^-(aq) + Zn^+2(aq) ==> Zn(OH)2(s) which is the net ionic equation.
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