What is the amount of Ca3(PO4)2 that can be prepared frim a mixture of 9g of Ca(OH)2 and 11g of H3PO4?
3Ca(OH)2 + 2H3PO4 >>>> Ca3(PO4)2 + 6H2O
First, we need to determine which reactant is limiting in the reaction.
First, calculate the molar mass of each reactant:
- Ca(OH)2: 1 mol Ca(OH)2 = 74.093 g/mol
- H3PO4: 1 mol H3PO4 = 98.0 g/mol
Next, calculate the number of moles of each reactant:
- Moles of Ca(OH)2 = 9g / 74.093 g/mol = 0.1213 mol
- Moles of H3PO4 = 11g / 98.0 g/mol = 0.1122 mol
Now, determine the limiting reactant by comparing the moles of each reactant to the stoichiometry of the reaction:
- From the balanced equation, the ratio of Ca(OH)2 to H3PO4 is 3:2.
- Since the molar ratio of Ca(OH)2 to H3PO4 is 3:2, there are more moles of Ca(OH)2 present, making H3PO4 the limiting reactant.
Now, calculate the theoretical yield of Ca3(PO4)2:
- Using the limiting reactant (H3PO4):
Moles of Ca3(PO4)2 = (0.1122 mol H3PO4) / (2 mol H3PO4) * (1 mol Ca3(PO4)2 / 3 mol H3PO4) = 0.0374 mol Ca3(PO4)2
- Convert moles to grams using the molar mass of Ca3(PO4)2:
Mass of Ca3(PO4)2 = 0.0374 mol * 310.177 g/mol = 11.61 g
Therefore, the amount of Ca3(PO4)2 that can be prepared from a mixture of 9g of Ca(OH)2 and 11g of H3PO4 is 11.61 g.