Help me answer this???

Question

You have reacted 100.00 mL of 0.949M calcium nitrate solution with 100.00 mL of 0.955M sodium phosphate solution. After filtering and drying the precipitate, you find that you have 8.630g of calcium phosphate solid. What is your percent yield? Report your answer to 1 decimal place. What would be the unit for the answer?



Answer 1

Step 1, balanced reaction:

3Ca(NO3)2 + 2Na3PO4 --> Ca3(PO4)2 + 6NaNO3

Step 2, calculate moles reactants:

Ca(NO3)2: 0.1L*0.949M = 0.0949 mol

Na3PO4: 0.1L*0.955M = 0.0955 mol

Step 3, determine limiting reagent:

You need Ca(NO3)2 in a ratio of 3:2 with Na3PO4, but you actually have it in the ratio of 0.0949:0.0955, clearly showing that Ca(NO3)2 is in short supply and is thus the limiting reagent. All calculations will be based on the amount of this reactant.

Step 4, determine how much Ca3(PO4)2 we should get in theory, assuming 100% yield:

0.0949 mol Ca(NO3)2 x 1 mol Ca3(PO4)2 per 3 mol Ca(NO3)2 = 0.0316 mol Ca3(PO4)2

Step 5, convert the theoretical yield of Ca3(PO4)2 to grams using molar mass of Ca3(PO4)2:

0.0316 mol x 310.174 g/mol = 9.801 g Ca3(PO4)2 theoretical

Step 6, find percent yield:

(actual divided by theoretical)*100 = (8.63/9.801)*100 = 88.0% yield