<p>Yes, you do need to balance them, if I remember correctly.
and no idea on the labs. my school took my lab fee and said “F U” basically.</p>
<p>Ha ha ha, that sucks.
I really have no idea either though, because my teacher hasn’t taught us sh|t. Only about 40 people at my school are taking the exam and we’re all screwed.</p>
<p>Okay, I have a reaction I have no idea what to do with.
Iron(II) chloride reacts with Gold(III) nitrate. What’ll be the net ionic equation? (Balance it).</p>
<p>Au3+ + 3Cl- –> AuCl3</p>
<p>?</p>
<p>I’m thinking Fe(NO3)2 is soluble…</p>
<p>It’s just a double replacement and if I did this correctly the balanced net ionic is essentially NORXN because all products and reactants are soluble in water. In any case the initial products are 3Fe(NO3)2 and 2AuCl3</p>
<p>i think AuCl3 forms a preciptate…</p>
<p>sitarhero, AuCl3 is very soluble in water.</p>
<p>Okay, that’s exactly what I thought ^
but apparently the answer says that since it’s a redox reaction, the net ionic equation is:
3Fe(2+)+Au(3+) —> 3Fe(3+)+Au(0)</p>
<p>Ahhh I gotcha! Ok with redox like that you initially cross out those “common spectator” ions (nitrates, potassium ions, clorides, etc) and you redox them (Them being the stuff left over). I don’t think I would have gotten that if you had pointed that out because I’m used to them saying acidified and you do a “real” redox or having it “just work out”.</p>
<p>^ I don’t understand. :(</p>
<p>55/75… good enough for a five, assuming decent FRQs?</p>
<p>Guesses for the equilibrium problem? I’m thinking titration, second guess Ksp.</p>
<p>And Serafina, if 70%/70% is a 5, 80%/60% would be a 5 too, right? The MC and FR have equal weightings?</p>
<p>Ok so I have FeCl2 and Au(NO3)3 right? I know those are both soluble so before the reaction I say Fe 2+ + 2Cl- + Au 3+ + 3(NO3)- . Right? So I know that the chloride ion and the nitrate ion will be my spectators (i’m pretty sure transition metals are more likely to redox). Also you know that anything that cloride or nitrate forms with will be soluble (mostly for chloride anyway) so that’s another way of looking at it. Do you see what I’m saying?</p>
<p>Okay I think I do. Thanks.
But why does the final equation have a coefficient of 3 in front of the Fe’s and no coefficient of 2 in front of the Au’s?
And also, how would you be able to do this type of thing without a reduction potential chart? You wouldn’t right?</p>
<p>Well the nice thing about AP chem is that all reactions go (in the reaction section) (yay!). Well for redox, the charges must balance on both sides. So what I would do is ignore the double-replacement type reaction coefficients and look at the charges and balance by inspection that way. And, as you know, each element must have the same number on each side. So those are two things for redox: charge and element balance.</p>
<p>OMG, just got it. Best feeling in the world.
Thanks, Terrence. Life saver. Now I can call my friend and take credit for this to make myself feel smart. :D</p>
<p>Is it just me or does Barrons Chem have a ****load of errors in it.</p>
<p>standard conditions = 25 C, 1 atm</p>
<p>correct?</p>
<p>^ Yes, except for gases. In the case of gases STP is 0 C, 1 atm.</p>
<p>^That’s standard temperature and pressure though?</p>
<p>Our chem teacher always taught us that standard conditions were 25 C and 1 atm, while STP was 0 C and 1 atm.</p>
<p>Apparently they’re two separate terms?</p>
<p>Haha go for it!</p>
<p>P.S. Luminouzz is correct</p>