# AP Chem Problem

<p>I have a 1st time AP teacher and he gave us this problem.....</p>

<p>Ti (3+) + HBrO --> TiO (2+) + Br (1-)</p>

<p>Question: Write the correctly balanced half reactions and net ionic equation for the skeletal equation shown above.</p>

<p>I did the balanced half reactions and got:
H2O + Ti (3+) --> TiO (2+) + 2H (1+) oxidation
2e- + H(1+) + HBrO --> Br(1-) + H2O reduction
How can I know where to put the (aq), (s) ??? </p>

<p>and the net ionic equation....??</p>

<p>You forgot the electron in the oxidation</p>

<p>H2O + Ti (3+) --> TiO (2+) + 2H (1+) + 1e- oxidation
2e- + H(1+) + HBrO --> Br(1-) + H2O reduction</p>

<p>2( H2O + Ti (3+) --> TiO (2+) + 2H (1+) )
1(2e- + H(1+) + HBrO --> Br(1-) + H2O )</p>

<p>2 H2O + 2 Ti (3+) --> 2 TiO (2+) + 4H (1+)
2e- + H(1+) + HBrO --> Br(1-) + H2O</p>

<p>H20 + 2 Ti (3+) + HBrO --> 2 TiO (2+) + Br(1-) + 3 H (1+)</p>

<p>I'm pretty sure that's it.</p>

<p>sonar, how do u kno where to put the (aq) and (s) and etc? do we have to do that for the AP exam?</p>

<p>You don't need state symbols on the AP, if I remember correctly.</p>

<p>But, it is helpful to know them on the net ionic equation free response section. In all honesty, the equation section was the one I did the worst on, I could never remember the general rules :P.</p>

<p>I can ratify that you don't need to label the state.</p>

<p>Thanks...cause my teacher told me we were suppose to put the state....:( this is his first year so I dont blame him....this is his first year teaching any AP or any chemistry class...</p>

<p>You need to memorize the following by the exam time for general solubility labeling (although for redox you probably can get away without labeling them). These don't cover every case, but they do cover a very good majority.</p>

<h1>General Solubility Rules</h1>

<h2>Mainly water soluble</h2>

<p>All nitrates are soluble.
All acetates are soluble.
All chlorides are soluble except AgCl, Hg2Cl2, and PbCl2
All bromides are soluble except AgBr, Hg2Br2, PbBr2, and HgBr2
All iodides are soluble except AgI, Hg2I2, PbI2, and HgI2
All sulfates are soluble except CaSO4, SrSO4, BaSO4, PbSO4, Hg2SO4, and Ag2SO4</p>

<h2>Mainly water insoluble</h2>

<p>All sulfides are insoluble except for those of 1A and 2A elements and (NH4)2)S
All carbonates are insoluble except for those of 1A and (NH4)2CO3
All phosphates are insoluble except for those of 1A and (NH4)3PO4
All hydroxides are insoluble except for those of 1A, Ba(OH)2, Sr(OH)2, and Ca(OH)2</p>

<p>Chem people....I need help...</p>

<p>The molecular formula of a compound containing only carbon and hydrogen is to be determined by analyzing its combustion products.
a) When burned completely, it produced 7.2 g of water and 14.15 g of CO2. What is the empirical formula of the original compound?
(What I did: I found the moles for each then divided both things CO2 and H2O by .4 (since that was the smalles mole) and I put CO2 + H2O as an answer) what do you think?
b) How many grams of oxygen were used in the combustion?
I have no idea.....</p>

<p>No, a good teacher will require state symbols. For the AP however, you don't need state symbols.</p>

<hr>

<p>no, you need to find the moles of carbon and the moles of hydrogen, not the moles of the molecules, then you're finding out how much there is of oxygen too.</p>

<p>7.2 g H2O * (1 mol H2O / 18 g H2O) * (2 mol H+ / 1 mol H2O) = .8 mol H
14.15 g CO2 * (1 mol CO2 / 44 g CO2) * (1 mol C / 1 mol CO2) = 0.321590 mol C</p>

<p>.8/0.321590 = 2.4876325088339222614840989399293
0.321590/0.321590 = 1</p>

<p>Since I used limited digits, I can assume that the ratio is 1:2.5 = 2:5</p>

<p>So, the empirical formula should be: C2H5</p>

<hr>

<p>grams of O2 is the easiest, a few ways you can do it, one way is</p>

<p>7.2 g H2O * (1 mol H2O / 18 g H2O) * (1 mol O / 1 H2O) * (16 g O / 1 mol O) = 6.4 g O
14.15 g CO2 * (1 mol CO2 / 44 g CO2) * (2 mol O / 1 mol CO2) * (16 g O / 1 mol CO2) = 10.2909 g O</p>

<p>6.4 g O + 10.2909 g O = 16.6909 = 17 g O (sig digits (7.2 g water))</p>

<p>You could also use the empirical formula to come up with a balanced equation and do the mole ratios, seeing as it's a hydrocarbon being burned (always equals CO2 + H2O).</p>

<p>C2H5 + O2 = CO2 +H2O</p>

<p>4 C2H5 + 13 O2 = 8 CO2 + 10 H2O</p>

<p>Then use either of the products and do stoichiometry.</p>

<p>7.2 g H2O * (1 mol H2O / 18 g H2O) * (13 mol O2 / 10 mol H2O) * (32 g O2 / 1 mol O2) = 16.64 g = 17 g O (sig digits)</p>

<p>14.15 CO2 * (1 mol CO2 / 44 g CO2) * (13 mol O2 / 8 mol CO2) * (32 g O2 / 1 mol O2) = 16.722727 = 17 g O (sig digits)</p>

<p>I think that's right. We didn't have to worry about a limiting reactant here because we had the products, I believe.</p>

<hr>

<p>As for solubility rules, if your teacher doesn't talk about them, look at the Princeton Review list, that's what you need to know. My solubility chart was my crutch last year :p.</p>

<p>Hey, thanks sonar!!! You are a genious!!! LOL really thank you...but I think you made a minor mistake....CO2 is actually 34 g. not 44 g. but I got the error....
element = number that you have * weight
C=1<em>12=12
O=2</em>16=32<br>
32+12= 34 grams....</p>

<p>32 + 12 = 34?</p>

<p>My algebra could maybe be rusty, but...???
Did I miss something?</p>

<p>No, CO2 is 44 g, but I remember it better as 44.0098. My teacher made us do 3 decimal points for all the elements, and 4 decimal places for oxygen.</p>

<p>Also, the real reason why there is no limiting reactant (I'm an idiot :p):</p>

<hr>

<p>There are only two reactants, the oxygen is in abundance, so obviously the hydrocarbon is the limiting reactant. Thus, only so much oxygen can be used.</p>

<p>Whoops! I made a big mistake!</p>

<p>You shouldn't do the second way (using the balanced equation). You're supposed to do it the first way. The second way only works because we're finding out how much oxygen there was, not how much of the hydrocarbon there was, since the ratios would be the same for O2 : products.</p>