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reddnwaiting
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Help please


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!!!!!
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ChenBeier
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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.
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reddnwaiting
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Thank you


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?
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ChenBeier
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It is still wrong.

Try to balance first Sb3+ with S2-

What do you get?
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reddnwaiting
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Think I got it!


I think I have it LOL

SbOH3 + Na2S-----Sb + NaOH
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ChenBeier
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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


Last edited by ChenBeier on Sun Sep 09, 2018 12:23 pm; edited 1 time in total
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reddnwaiting
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Thank you


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

Thank you for your help!
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