Final answer:
CaCl2 is the limiting reactant, and 213.9 grams of Ca3(PO4)2 are theoretically produced. If 150 grams are actually recovered, the percent yield is 70.13%.
Step-by-step explanation:
To determine which reactant is limiting in the reaction of Na3PO4 and CaCl2, we need to consider the stoichiometry of the balanced chemical equation provided. According to the equation, 2 moles of Na3PO4 react with 3 moles of CaCl2 to produce 1 mole of Ca3(PO4)2 and 6 moles of NaCl.
First, we convert the masses to moles using the molar masses: Na3PO4 (163.94 g/mol) and CaCl2 (110.98 g/mol):
- For Na3PO4: 307 g / 163.94 g/mol = 1.873 moles
- For CaCl2: 227 g / 110.98 g/mol = 2.045 moles
The mole ratio of Na3PO4 to CaCl2 is 2:3, or 1:1.5.
Therefore, for 2 moles of Na3PO4, we need 3 moles of CaCl2. The moles of Na3PO4 multiplied by 1.5 equals 2.810 moles, which is more than the available 2.045 moles of CaCl2. So, CaCl2 is the limiting reactant.
Using the stoichiometry of the balanced reaction, we can find out how many grams of Ca3(PO4)2 are produced. Since CaCl2 is the limiting reactant, we base our calculation on its moles:
2.045 moles CaCl2 × (1 mole Ca3(PO4)2 / 3 moles CaCl2) × (310.18 g/mol Ca3(PO4)2) = 213.9 g Ca3(PO4)2
To calculate the percent yield, we divide the actual yield by the theoretical yield and multiply by 100:
(150 g / 213.9 g) × 100 = 70.13%
The percent yield of the reaction is 70.13%.