Asked by Jennie
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).
b.What is the actual yield of Ti(s) in this experiment?
c.Calculate the % yield of Ti(s) in this experiment.
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).
b.What is the actual yield of Ti(s) in this experiment?
c.Calculate the % yield of Ti(s) in this experiment.
Answers
Answered by
DrBob222
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.
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.
Answered by
Jennie
Thanks DrBob222!!! You have helped me so much!!
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