Stoichiometrics:
a 36 grams sample of calcium hydroxide is allowed to react with 40.5 grams of phosphoric acid according to the following reaction.
3Ca(OH)2 + 2H3PO4 = Ca3(PO4)2 + 6H2O
how many grams of calcium phosphate are produced?
how many moles of Calcium phosphate are produced?
how many moles of excess reagent remain?
if 43.5 grams of calcium phosphate are produced what is the % yield?
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%.
To determine the grams of calcium phosphate produced, we need to first find the limiting reactant. The limiting reactant is the reactant that will be completely consumed in the reaction, thus determining the amount of product that can be formed.
Step 1: Calculate the number of moles of each reactant:
Mass of Ca(OH)2 = 36 grams
Molar mass of Ca(OH)2 = 40.08 g/mol (calcium, 1 atom) + 2 * (16.00 g/mol (oxygen, 2 atoms) + 1.01 g/mol (hydrogen, 2 atoms)) = 74.10 g/mol
Number of moles of Ca(OH)2 = Mass / Molar mass = 36 g / 74.10 g/mol = 0.486 moles
Mass of H3PO4 = 40.5 grams
Molar mass of H3PO4 = 3 * (1.01 g/mol (hydrogen, 3 atoms)) + 1.01 g/mol (phosphorus) + 4 * (16.00 g/mol (oxygen, 4 atoms)) = 98.00 g/mol
Number of moles of H3PO4 = Mass / Molar mass = 40.5 g / 98.00 g/mol = 0.413 moles
Step 2: Determine the limiting reactant:
To do this, we compare the stoichiometric coefficients of the reactants in the balanced equation.
From the balanced equation: 3 moles of Ca(OH)2 react with 2 moles of H3PO4 to produce 1 mole of Ca3(PO4)2.
So, the ratio of moles of Ca(OH)2 to H3PO4 is 3/2.
If we convert the moles of H3PO4 to the same ratio as Ca(OH)2, we get:
Number of moles of H3PO4 * (3/2) = 0.413 moles * (3/2) = 0.6195 moles
Since 0.486 moles of Ca(OH)2 is less than 0.6195 moles of H3PO4, Ca(OH)2 is the limiting reactant.
Step 3: Calculate the moles of Ca3(PO4)2 produced:
From the balanced equation: 3 moles of Ca(OH)2 react with 1 mole of Ca3(PO4)2.
So, the mole ratio of Ca(OH)2 to Ca3(PO4)2 is 3/1.
Number of moles of Ca3(PO4)2 = Number of moles of Ca(OH)2 = 0.486 moles
Step 4: Calculate the grams of Ca3(PO4)2 produced:
Molar mass of Ca3(PO4)2 = 3 * (40.08 g/mol (calcium, 3 atoms)) + 2 * (31.00 g/mol (phosphorus, 2 atoms)) + 8 * (16.00 g/mol (oxygen, 8 atoms)) = 310.18 g/mol
Mass of Ca3(PO4)2 = Number of moles * Molar mass = 0.486 moles * 310.18 g/mol = 150.79 grams
Therefore, 150.79 grams of calcium phosphate are produced.
To find the moles of calcium phosphate produced, we can use the same method as above but using the molar mass of Ca3(PO4)2:
Number of moles of Ca3(PO4)2 = Mass / Molar mass = 150.79 grams / 310.18 g/mol = 0.486 moles
Thus, 0.486 moles of calcium phosphate are produced.
To find the moles of excess reagent (in this case, the H3PO4) remaining, we need to calculate the moles of the limiting reactant used:
From the balanced equation: 3 moles of Ca(OH)2 react with 2 moles of H3PO4.
So, the mole ratio of Ca(OH)2 to H3PO4 is 3/2.
Number of moles of H3PO4 reacted = Number of moles of Ca(OH)2 used * (2/3) = 0.486 moles * (2/3) = 0.324 moles
To find the moles of excess reagent remaining, we subtract the moles reacted from the original moles of the reagent:
Moles of excess H3PO4 remaining = 0.413 moles - 0.324 moles = 0.089 moles
If 43.5 grams of calcium phosphate are produced, we can calculate the percent yield by comparing the actual yield to the theoretical yield:
Theoretical yield = 0.486 moles * 310.18 g/mol = 150.79 grams
Percent yield = (Actual yield / Theoretical yield) * 100% = (43.5 grams / 150.79 grams) * 100% = 28.85%