Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Kax0r on November 11, 2010, 12:37:48 PM
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13) Write a balanced net ionic equation for the reaction of Pb(NO3)2(aq) with NaI(aq).
A) Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 I-(aq) PbI2(s) + 2 Na+(aq) + 2 NO3-(aq)
B) Pb(NO3)2(aq) + 2 NaI(aq) PbI2(s) + 2 NaNO3(aq)
C) Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 I-(aq) Pb2+(aq) + 2 I-(aq) + 2 Na+(aq) + 2 NO3-(aq)
D) Pb2+(aq) + 2 I-(aq) PbI2(s)
15) Write a balanced net ionic equation for the reaction of NiBr2(aq) with (NH4)2S(aq).
A) Ni2+(aq) + 2 Br-(aq) + 2 NH4+(aq) + S2-(aq) NiS(s) + 2 NH4Br(s)
B) Ni2+(aq) + S2-(aq) NiS(s)
C) NiBr2(aq) + (NH4)2S(aq) NiS(s) + 2 NH4Br(aq)
D) Ni2+(aq) + 2 Br-(aq) + 2 NH4+(aq) + S2-(aq) NiS(s) + 2 NH4+(aq) + 2 Br-(aq)
Okay, I've gotten several answers, a couple say D first B second, but these threads seem to disagree:
http://answers.yahoo.com/question/index?…
http://jchemed.chem.wisc.edu/JCESoft/CCA…
Who is right? I'm so confused :(
Edit: I'm pretty confident in D and B but I wanted to make sure. It looks like the full ionic equation for the first one would be:
Pb2+2NO3+2Na+2I = PbI2 + 2Na + 2 NO3.
which, when removing spectator ions would be- Pb2+ 2I = PbI2.
For the second one, it would be much the same. Ni+2Br+2NH4+S=NiS+ 2 NH4 + 2Br, removing spec ions, Ni+S=NiS
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yeah look this is due in like a half hour so can I please get some answers?
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Try here: forum rules (http://www.chemicalforums.com/index.php?topic=33740.0).