Posted by Bill on Monday, May 21, 2012 at 7:40pm.
I see you've posted this a number of times. What is it you don't understand? Perhaps I can get you started.
This is a few of my final review questions. I basically just need to know the steps for each one because my teacher doesn't explain them very well. I would really appreciate some help.
Please I really need your help here, you seem to be very knowledgable in the subject.
Is #1 Al bicarbonate? You need to write the formula or the full name; otherwise it doesn't make sense. I'll assume Al(HCO3)3. I'll call tartaric acid H2T.
2Al(HCO3)3 + 3H2T ==> Al2T3 + 6CO2 + 6H2O
How many mols CO2 were formed? Use PV - nRT, substitute the conditions and solve for n = number of mols. Convert mols CO2 to mols Al(HCO3)3 by using the coefficients in the balanced equation. The M bicarbonate = mols/L. You have mols and M, solve for L. (Note: you have no volume given for H2T even though that appears to the the limiting reagent from the way the problem iks stated.)
2.
2. 3.1 x 10^23 molecules of .0035 M hydriodic acid reacts with 25 grams of magnesium bicarbonate how much carbon dioxide if at stp?
This is a limiting reagent problem.
Mg(HCO3)2 + 2HI ==> MgI2 + 2H2O + 2CO2
mols HI = #molecules/6.02E23 = ?
mols Mg(HCO3)2 = grams/molar mass.
Using the coefficients in the balanced equation convert mols HI to mols of the product.
Do the same for mols Mg(HCO3)2.
It is likely that the two answers for mols of the product will not agree; therefore, one of them must be wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent. The problem doesn't tell us the units so mols may be sufficient. If you want grams, that is mols x molar mass = grams. If you want volume, that is mols x (22.4L/mol) = L CO2.
#3. What's the significance of T1, T2, T3? You don't have units on the other numbers
mols NaOH = M x L = ?
(NaOH) = (OH^-) = moles/L soln
pOH = -log(OH^-)
For pH,
pH + pOH = pKw = 14. You know pOH and pKw, solve for pH.
Thank you very much!
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