Topic: Balancing Chemical Equations (Read 1721 times)

Sibbby2001 · « **on:** April 16, 2021, 12:54:42 PM »

FeCl3.6H2O + Ph2Hacac + NaOH = (FeIII(Ph2acac)3) + NaCl + H2O

Please can anyone help me balance this equation? Just note the 3 I's after Fe is the charge.

Borek · « **Reply #1 on:** April 16, 2021, 01:05:24 PM »

What is Ph2Hacac intended to be? Is the disappearing H an acidic proton, or a typo?

Sibbby2001 · « **Reply #2 on:** April 16, 2021, 01:12:22 PM »

The H in the (Ph2Hacac) is this 1,3-diphenyl-1,3-propanedione in its enol form.

Sibbby2001 · « **Reply #3 on:** April 16, 2021, 01:19:25 PM »

The ligand is abbreviated as Ph2Hacac with the deprotonated anionic form called Ph2acac

I've managed to get
FeCl3.6H2O + 3Ph2Hacac + 3NaOH = (FeIII(Ph2acac)3) + 3NaCl + 9H2O as my
answer. I don't think it's right though.

Borek · « **Reply #4 on:** April 16, 2021, 02:01:54 PM »

Looks OK to me.

FeCl₃·6H₂O + 3Ph₂Hacac + 3NaOH

Fe(Ph₂acac)₃ + 3NaCl + 9H₂O

Sibbby2001 · « **Reply #5 on:** April 16, 2021, 02:27:09 PM »

Thank you, Borek. I was told that it was incorrect. So I was looking over it to find an error that wasn't there.

Topic: Balancing Chemical Equations (Read 1721 times)

Sibbby2001

Balancing Chemical Equations

Borek

Re: Balancing Chemical Equations

Sibbby2001

Re: Balancing Chemical Equations

Sibbby2001

Re: Balancing Chemical Equations

Borek

Re: Balancing Chemical Equations

Sibbby2001

Re: Balancing Chemical Equations

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