Solve Net Ionic Equations: Chromium, Aluminium, Zinc

[joejo]

hi, I'm studying for the chemistry SAT and have a question that I'm not sure that I've done right...can someone please check my answer...thanks in advance!

Write a net ionic equation for the following:

a) Chromium dipped into silver nitrate
b) Aluminmum dropped into a bath of sulphuric acid
c) Zinc dipped into a solution of led (II) nitrate

answers:

a) Cr(s) + Ag+ (aq) --> Cr2+ (aq) + 2Ag (s)
b) d)Al3+ + H2SO4 --> Al2(So4)3 + H2
c) Zn2+ + Pb(NO3)2 --> Zn(NO3)2 + Pb2+

im not sure if I have to have the (s), (aq) etc...

can someone correct my answers if they are wrong...

thanks again

 

[The Bob]

A looks right. You have state symbols missing from B and in an ionic equation the hydrogen ions are not changing so are not needed (I believe). C also has no symbols and again the lead is not really changing for an ionic equation.

The Bob (2004 ©)

 

[joejo]

my friend helped me get a...the answers for b and c are mine...I still don't get it...can you please show me?

 

[joejo]

someone please my test is on thursday i need to study! Thanks guys

 

[dextercioby]

2\mbox{Al}_{(\mbox{s})}+6\mbox{H}^{+}_{(\mbox{aq})} \rightarrow 2\mbox{Al}^{3+}_{(\mbox{aq})}+3\mbox{H}_{2}\uparrow

(sulphate ions are spectators)

\mbox{Zn}_{(\mbox{s})}+\mbox{Pb}^{2+}_{(\mbox{aq})}\rightarrow \mbox{Zn}^{2+}_{(\mbox{aq})} +\mbox{Pb}\downarrow

(nitrate ions are spectators)


Daniel.

 

[dextercioby]

And the first one needs another 2.It should read

\mbox{Cr}_{(\mbox{s})}+2\mbox{Ag}^{+}_{(\mbox{aq})}\rightarrow \mbox{Cr}^{2+}_{(\mbox{aq})}+2\mbox{Ag}\downarrow

(Nitrate ions are spectators)


Daniel.

 

[whozum]

b. aluminum into sulfuric acid
If I remember how to do these correctly:
2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2

Break it down into its compounds:

2Al + 3H_2^+ + 3SO_4^{2-} \rightarrow Al_2^{3+} + (SO_4)_3^{2-} + 3H_2

Look on both sides to see if anything is idle during the process. It looks like everythings reacting, even the hydrogens look like theyre doing something, since they go from H_2^+ \rightarrow H_2.

I would go to http://www.towson.edu/~ladon/netionic.html for some more help.

edit: Sulfate is spectating as dex said, i didnt see the coefficient. Cros it out

 

[dextercioby]

The net charge in your equation,in the form written,doesn't add to zero in any of the 2 members,though it should...U have 6 protons from 3 sulphuric acid molecules...They're not 3 molecule-ion of Hydrogen...They have total charge "6+",and not "3+" as your notation would assume...

Daniel.

 

[whozum]

So in this case you would write H as a single atom and not a diatomic? theyre teaching us to always write the diatomics as diatomics. My notation is probably unorthodox but when I counted it it added up right.

2Al + 3H_2^+ + 3SO_4^{2-} \rightarrow Al_2^{3+} + (SO_4)_3^{2-} + 3H_2

Left:
2Al = 0
3H_2 += +6
3SO_4 2- = -6

Right:
Al_2 3+= 6+
SO_4 _3 2- = 6-
H_2 = 0

 

[dextercioby]

That is not diatomic (in the LHS),it's simply 2 \mbox{H}^{+} ions (protons) involved in a chemical bound (covalent,but here we're interested in ionic reactions,though the chemical bounds are not necessarily ionic) with 2 atoms of oxygen from the 4 involved in the sulphate ion...

Notation is essential only when it's wrong.

Daniel.

 

[whozum]

Ok I see where I went wrong, thanks. Learn from my mistakes joejo :)

 

[joejo]

no u've totally lost me whos right...whozum or dexter...

dexter your right, right??

 

[dextercioby]

>50% of the time,this one included

Daniel.

 

[joejo]

thanks guys...


i don't get u dexter...are you saying ur guessing?

 

[whozum]

dexters always right.
he has a phd in everything.

 

[joejo]

thank a lot guys