Dr. Bob222, Yesterday I posted a question and you answered me with how to calculate it.. Can you tell me if I'm doing this correctly?? Please!
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. Calculate the theoretical yield of Ti(s).
Ok, The atomic weight of Ti=47.88 Cl=35.45 (*4) Mg=24.31 (*2)
Total Molar Mass of TiCl4=189.68g/mol
to calculate TY of Ti = 47.88/189.68= 0.252 moles of Ti??? Where you wrote TiCl4 * (1mol Ti/1mol TiCl4) am I supposed to multiply 189.68 *0.252?? if so that would = 47.80 ???
b.What is the actual yield of Ti(s) in this experiment? AY =OHH this is given already, as 7.91*10E6 ok.. So then...
c.Calculate the % yield of Ti(s) in this experiment. AY/TY *100 which would be 7.91*10E6/??? I can't figure out the correct theoretical yield.. I think I'm confusing myself... Help!!
Chemistry - DrBob222, Tuesday, February 24, 2015 at 11:36pm
Remember how to work this type stoichiometry problem.
Step 1. Convert what you have (in this case TiCl4) to mols. mols = grams/molar mass
Step 2. Using the coefficients in the balanced equation, convert mols of what you have to mols of what you want(in this case Ti). With the coefficients you can convert anything to anything.
mols TiCl4 x (1 mol Ti/1 mol TiCl4) = mols TiCl4 x 1/1 = ?
Step 3. Now convert mols of what you have to grams. grams = mols Ti x atomic mass Ti. This is the theoretical yield (i.e., as if it were 100%). Let's call this TY for theoretical yield.
Step 4. It isn't 100%. The actual yield from the problem is 7.91E6 g. Let's call this AY for actual yield.
Step 5. % yield = (AY/TY)*100 = ?
Print this out. This five step procedure will work all of the stoichiometry problems (limiting reagent problems take an extra step) and if the problem is solution mols = M x L.
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