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