Superphosphate, a water soluble fertilizer, is sometimes marked as a "triple phosphate". It is a mixture of Ca(H2PO4)2 + 2CaSO4.
Ca3(PO4)2 + 2H2SO4 --> Ca(H2PO4)2 + 2CaSO4
If you treat 400g of calcium phosphate with 267g of sulfuric acid, how many g of superphosphate can be formed?
My question is, if this entire thing is superphosphate --> Ca(H2PO4)2 + 2CaSO4, do I have to add the mass of both Ca(H2PO4)2 and 2CaSO4? I've found the mass of Ca(H2PO4)2 to be 300g.
Yes. You find the mass of Ca(H2PO4)2 and add that to the mass of CaSO4 to find the mass of the "superphosphate".
To find out how many grams of superphosphate can be formed, we need to determine the limiting reactant in the reaction between calcium phosphate and sulfuric acid. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed.
First, let's calculate the grams of calcium sulfate (CaSO4) formed:
Molar mass of CaSO4 = 40.08 g/mol (Ca) + 32.07 g/mol (S) + 4 * 16.00 g/mol (O) = 136.14 g/mol
The molar ratio between calcium phosphate (Ca3(PO4)2) and calcium sulfate (CaSO4) in the balanced chemical equation is 2:2, meaning that 2 moles of calcium phosphate react to produce 2 moles of calcium sulfate.
To calculate the moles of calcium sulfate formed, we use the given mass of calcium phosphate and its molar mass:
Moles of calcium phosphate = 400 g / (3 * 40.08 g/mol + 2 * 31.03 g/mol + 8 * 16.00 g/mol) ≈ 1.07 mol
Since the molar ratio is 2:2, the moles of calcium sulfate formed will also be 1.07 mol.
To calculate the grams of calcium sulfate, we multiply the moles by the molar mass:
Mass of calcium sulfate = 1.07 mol * 136.14 g/mol ≈ 145.72 g
Next, let's calculate the grams of calcium dihydrogen phosphate (Ca(H2PO4)2) formed:
The molar mass of Ca(H2PO4)2 = 40.08 g/mol (Ca) + 2 * 1.01 g/mol (H) + 2 * 31.03 g/mol (P) + 8 * 16.00 g/mol (O) = 234.05 g/mol
The molar ratio between calcium phosphate (Ca3(PO4)2) and calcium dihydrogen phosphate (Ca(H2PO4)2) in the balanced chemical equation is 2:1, meaning that 2 moles of calcium phosphate react to produce 1 mole of calcium dihydrogen phosphate.
To calculate the moles of calcium dihydrogen phosphate formed, we use the given mass of calcium phosphate and its molar mass:
Moles of calcium dihydrogen phosphate = (400 g / (3 * 40.08 g/mol + 2 * 31.03 g/mol + 8 * 16.00 g/mol)) / 2 ≈ 0.535 mol
The moles of calcium dihydrogen phosphate formed will be half of the moles of calcium phosphate since the molar ratio is 2:1.
To calculate the grams of calcium dihydrogen phosphate, we multiply the moles by the molar mass:
Mass of calcium dihydrogen phosphate = 0.535 mol * 234.05 g/mol ≈ 124.79 g
Thus, the total mass of superphosphate (Ca(H2PO4)2 + 2CaSO4) formed will be the sum of the masses of calcium dihydrogen phosphate (Ca(H2PO4)2) and calcium sulfate (CaSO4):
Total mass of superphosphate = Mass of calcium dihydrogen phosphate + Mass of calcium sulfate
= 124.79 g + 145.72 g
= 270.51 g
Therefore, approximately 270.51 grams of superphosphate can be formed.
To find the mass of superphosphate formed, you need to consider the stoichiometry of the reaction. The balanced equation shows that 1 mole of calcium phosphate (Ca3(PO4)2) reacts with 2 moles of sulfuric acid (H2SO4) to form 1 mole of calcium dihydrogen phosphate (Ca(H2PO4)2) and 2 moles of calcium sulfate (CaSO4).
First, you need to find the moles of calcium phosphate and sulfuric acid using their respective masses and molar masses:
Molar mass of calcium phosphate (Ca3(PO4)2) = 310.18 g/mol
Molar mass of sulfuric acid (H2SO4) = 98.09 g/mol
Moles of calcium phosphate (Ca3(PO4)2) = Mass of calcium phosphate / Molar mass of calcium phosphate
Moles of sulfuric acid (H2SO4) = Mass of sulfuric acid / Molar mass of sulfuric acid
Moles of calcium phosphate = 400 g / 310.18 g/mol
Moles of sulfuric acid = 267 g / 98.09 g/mol
Now, you need to compare the stoichiometry of the reaction to determine the limiting reagent. The reactant that produces the lesser amount of product is the limiting reagent.
From the balanced equation, you can see that 1 mole of calcium phosphate reacts with 2 moles of sulfuric acid. Therefore, the moles of sulfuric acid required for the reaction is twice the moles of calcium phosphate.
Moles of sulfuric acid required = 2 * Moles of calcium phosphate
Moles of sulfuric acid required = 2 * (400 g / 310.18 g/mol)
Now compare the moles of sulfuric acid required with the moles of sulfuric acid available:
If the moles of sulfuric acid available is greater than the moles of sulfuric acid required, then sulfuric acid is in excess, and the moles of calcium phosphate will determine the moles of product formed.
If the moles of sulfuric acid required is greater than the moles of sulfuric acid available, then sulfuric acid is the limiting reagent, and the moles of sulfuric acid will determine the moles of product formed.
Finally, calculate the mass of superphosphate formed using the limiting reagent:
If the limiting reagent is calcium phosphate:
Mass of superphosphate formed = Moles of calcium phosphate * Molar mass of superphosphate
Mass of superphosphate formed = (400 g / 310.18 g/mol) * (molar mass of Ca(H2PO4)2)
If the limiting reagent is sulfuric acid:
Mass of superphosphate formed = Moles of sulfuric acid * Molar mass of superphosphate
Mass of superphosphate formed = (267 g / 98.09 g/mol) * (molar mass of Ca(H2PO4)2)
Note that the molar mass of Ca(H2PO4)2 was provided as 300 g/mol in the question, so you can substitute that value in the equations.
The mass of Ca(H2PO4)2 is not required in this calculation, and the mass of 2CaSO4 is not relevant to determining the mass of superphosphate formed in this specific reaction.