You have a limiting reagent problem here.
1.Convert grams to mols. Mols = g/molar mass.
2. Convert mols Ca to mols CaF2. Convert mols F2 to mols CaF2.
3. The smaller number of mols of CaF2 will be the one to pick and the reagent with that result will be the limiting reagent.
4. Using the smaller number of mols CaF2, convert to grams CaF2. That is the theoretical yield.
5.
%yield = [(actual yield/theoretical yield)]*100
Note that the acutal yield is 4.02 g.
Ca + F2 --> CaF2
The mass of CaF2 that results from the reaction of 3.00 g of calcium and 2.00 g of fluorine is 4.02 g. What is the percent yield? (Atomic weights: Ca = 40.08, F = 19.00).
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