To find the grams of calcium phosphate produced, we need to calculate the stoichiometric ratio of calcium hydroxide to calcium phosphate.
Step 1: Calculate the moles of calcium hydroxide and phosphoric acid
Molar mass of Ca(OH)2 = 40.08 g/mol + 2(16.00 g/mol) + 2(1.01 g/mol) = 74.10 g/mol
Moles of Ca(OH)2 = Mass / Molar mass = 36 g / 74.10 g/mol = 0.486 mol
Molar mass of H3PO4 = 1.01 g/mol + 3(1.01 g/mol) + 4(16.00 g/mol) = 98.00 g/mol
Moles of H3PO4 = Mass / Molar mass = 40.5 g / 98.00 g/mol = 0.413 mol
Step 2: Determine the limiting reagent
Since the stoichiometric ratio between Ca(OH)2 and H3PO4 is 3:2, we need to check which reagent is the limiting reagent.
Divide the number of moles of each reagent by its stoichiometric coefficient:
Moles of Ca(OH)2 / Stoichiometric coefficient of Ca(OH)2 = 0.486 mol / 3 = 0.162 mol
Moles of H3PO4 / Stoichiometric coefficient of H3PO4 = 0.413 mol / 2 = 0.207 mol
The smaller value, 0.162 mol, corresponds to Ca(OH)2. This means that Ca(OH)2 is the limiting reagent, and H3PO4 is in excess.
Step 3: Calculate the moles of calcium phosphate and excess reagent
From the balanced equation, we can see that the stoichiometric ratio between Ca(OH)2 and Ca3(PO4)2 is 3:1.
Moles of Ca3(PO4)2 = Moles of limiting reagent (Ca(OH)2) = 0.162 mol
To calculate the moles of excess reagent (H3PO4), we need to subtract the moles of H3PO4 used from the initial moles of H3PO4:
Moles of excess reagent (H3PO4) = Initial moles of H3PO4 - Moles used = 0.413 mol - 2(0.162 mol) = 0.089 mol
Step 4: Calculate the grams of calcium phosphate produced
Molar mass of Ca3(PO4)2 = 3(40.08 g/mol) + 2(30.97 g/mol) + 8(16.00 g/mol) = 310.18 g/mol
Grams of Ca3(PO4)2 = Moles of Ca3(PO4)2 * Molar mass = 0.162 mol * 310.18 g/mol = 50.32 g
Step 5: Calculate the % yield
Actual yield = 43.5 g
Theoretical yield = grams of Ca3(PO4)2 = 50.32 g
% yield = (Actual yield / Theoretical yield) * 100
% yield = (43.5 g / 50.32 g) * 100 = 86.33%
Therefore, the grams of calcium phosphate produced is 50.32 g, the moles of calcium phosphate produced is 0.162 mol, the moles of excess reagent remaining is 0.089 mol, and the percent yield is 86.33%.