Computing the molar mass? Using this equation for all questions: 3 CaI2 (aq) + 2

Question

Computing the molar mass?

Using this equation for all questions: 3 CaI2 (aq) + 2 (NH4)3PO4 (aq) --> Ca3(PO4)2 (s) + 6 NH4I (aq)

1. If you weigh out 12.43 g of CaI2 (calcium iodide), how many moles do you have?

2. What is the mole-mole ratio (conversion factor) for CaI2 (calcium iodide) to Ca3(PO4)2 (calcium phosphate)?

3. If you start with 16.35 g of CaI2, how much (in grams) solid Ca3(PO4)2 is expected (theoretical yield)?

4. If you run the experiment and isolate 4.77g of Ca3(PO4)2, what is your percent yield?

Explanation / Answer

1. mass of CaI2 taken = 12.43g

molar mass of CaI2 = 293.887g/mol

no. of moles = mass taken / molar mass= 12.43/293.887 = 0.0423moles.

2. every 3 moles of CaI2 is reacted to produce 1 mole of Ca3(PO4)2. So, 0.0423moles CaI2 will produce moles of Ca3(PO4)2 = 0.0423/3 = 0.0141moles.

Thus conversion factor = 1/3.

3. moles of CaI2 present = 16.35g/293.887g/mol = 0.0556moles

Moles of Ca3(PO4)2 expected = 0.0556moles / 3 = 0.0185moles

So, mass produced of Ca3(PO4)2= 0.0185 moles x 310.17g/mol (molar mass) = 5.572g

4.% yield = observed yield/theoretical yield x 100 = 4.77/5.752 x 100 = 82.93%

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