a. (10 points) Calculate the theoretical yield of Ti(s).
In order to find TY I have to find the limiting agent.. ?? ugh.. Ok, Ti=47.88 Cl4=35.45*4... To find moles in TiCl4 I take 3.54*10^7 * 1moleTiCl4/189.68 = 184,521.89 moles? But I need TY of Ti?? 7.91*10^6 * 1mole Ti/47.88=165,204.68 moles Ti???? I'm completely lost.. I guess I don't even know how to correctly find the moles of things?? I'm taking this class online at UTA with no help from a teacher and I'm trying to teach myself.. and I'm struggling so bad!!!! I think after your help I am doing it right then I confuse myself again..
This is not a limiting reagent problem. The amount of only one reactant is given so that is the limiting reagent. There is nothing else to consider. mols = grams/molar mass is how you determine mols.
mols TiCl4, the starting material, is 3.54E7grams TiCl4/molar mass TiCl4 = 3.54E7/189/68 = 1.87E5 mols TiCl4. Then you convert that to mols Ti to find how much Ti can be prepared.
1.87E5 mols TiCl4 x (1 mol Ti/1 mol TiCl4) = 1.87E5 x 1/1 = 1.87E5 mols Ti can be prepared. Then convert mols Ti to grams Ti by grams Ti x atomic mass Ti = 1.87E5 mols Ti x 47.88 = about 8.9E6 grams.
b. (5 points) What is the actual yield of Ti(s) in this experiment? 7.91*10^6 it's given...yes
c. (10 points) Calculate the % yield of Ti(s) in this experiment. Percent yield is AY/TY so would it be?? 7.91*10^6/165,204.68?? the moles of Ti? or is that the TY of Ti? I'm so confused.. You said before you got the TY to be 8.9E6? how??
3.54 x 10^7 g TiCl4 x (1 mol/189.68 g) x (1 mol Ti/1 mol TiCl4) x (47.88 g/1 mol) = 8.94 x 10^6 g Ti ????
Why do you keep trying to get that 165 whatever number into the mix. That came from the 7.91E6/189.68 but that is NOT the mols you stated with. 7.91E6 is the grams of Ti you ENDED with and doesn't have anything to do with how you started. You are following that 4/5 step procedure I gave you at the beginning. You're mixing starting and ending materials/products. So the theoreticfal yield is that 8.94E6 grams.
Percent yield is equal to the actual amount yield divided by the theoretical amount, so: 7.91 x 10^6 all divided by 8.94 x 10^6 = 88.5% yield ???
This is right
Dr. Bob222 Please don't get frustrated with me, I'm really trying to figure this out... UGH!!
I'm starting over again... Please advise...
. Consider the following chemical equation:
TiCl4(g) + 2Mg(l) → Ti(s) + 2MgCl2(l)
3.54×107g of TiCl4(g) were reacted completely and 7.91×106g of Ti(s) was obtained in the experiment.
a. (10 points) Calculate the theoretical yield of Ti(s).
In order to find TY I have to find the limiting agent.. ?? ugh.. Ok, Ti=47.88 Cl4=35.45*4... To find moles in TiCl4 I take 3.54*10^7 * 1moleTiCl4/189.68 = 184,521.89 moles? But I need TY of Ti?? 7.91*10^6 * 1mole Ti/47.88=165,204.68 moles Ti???? I'm completely lost.. I guess I don't even know how to correctly find the moles of things?? I'm taking this class online at UTA with no help from a teacher and I'm trying to teach myself.. and I'm struggling so bad!!!! I think after your help I am doing it right then I confuse myself again..
b. (5 points) What is the actual yield of Ti(s) in this experiment? 7.91*10^6 it's given...
c. (10 points) Calculate the % yield of Ti(s) in this experiment. Percent yield is AY/TY so would it be?? 7.91*10^6/165,204.68?? the moles of Ti? or is that the TY of Ti? I'm so confused.. You said before you got the TY to be 8.9E6? how??
3.54 x 10^7 g TiCl4 x (1 mol/189.68 g) x (1 mol Ti/1 mol TiCl4) x (47.88 g/1 mol) = 8.94 x 10^6 g Ti ????
Percent yield is equal to the actual amount yield divided by the theoretical amount, so: 7.91 x 10^6 all divided by 8.94 x 10^6 = 88.5% yield ???
Is this right???
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