Chemistry Forum

I am not sure if I have solved this correctly.

Sb (OH)₃ +Na₂S --- SbS₃

I have attempted to check my answer, but I can't find it online.

Thank you!!!!!

The equation is not balanced. On left side and right side the number of atoms must be equal. Check also oxidation number of stibium.

Try again.

The lab required Sb₃+

Sb OH₃ + Na₂ S-----2SbS₃ +2Na OH₃

I have a problem grasping is it the total number of charges that need to be equal on both sides?

It is still wrong.

Try to balance first Sb₃+ with S₂-

What do you get?

I think I have it LOL

SbOH₃ + Na₂S-----Sb + NaOH

No

In Sb₃+ + S₂- , we have 3+ and 2- . Like in mathematics 1/3 and 1/2 you have to go 6.

2 * 3+ = -(3* 2-)

2 Sb₃+ + 3 S₂- => Sb₂S₃

With your compounds it means

2 Sb(OH)₃ + 3 Na₂S => Sb₂S₃ + 6 NaOH

Thank you for your help. I see how you solved it and I will practice more.

Thank you for your help!

reddnwaiting wrote:I am not sure if I have solved this correctly.

Sb (OH)₃ +Na₂S --- SbS₃

I have attempted to check my answer, but I can't find it online.

Thank you!!!!!

2Sb(OH)₃(s)+3NA₂S(ac)=SB₂S₃(s)+6NaOH(ac)

-Sb has +3
-S has -2
-Na has+1
-OH has -1

The problem was solved already above.