Superphosphate, or sometimes called triple phosphate, is a commonly used water-soluble

Question

Superphosphate, or sometimes called triple phosphate, is a commonly used water-soluble

fertilizer. It is produced by the reaction:
𝐶𝑎3
(𝑃𝑂4
)2 + 𝐻2𝑆𝑂4 → 𝐶𝑎(𝐻2𝑃𝑂4
)2 + 𝐶𝑎𝑆𝑂4
If we mix 200.0 g of 𝐶𝑎3
(𝑃𝑂4
)2 with 133.5 g of 𝐻2𝑆𝑂4, how much 𝐶𝑎(𝐻2𝑃𝑂4
)2 (in grams)
can be produced?

Answer 1

I don't know why you chose to post your question in such a weird fashion. It makes it hard to understand, hard to read, and hard to follow. What you have here is a limiting reagent (LR) problem. You know you have a LR problem when amounts are given for BOTH reactants. I do these the long way.

..............𝐶𝑎3(𝑃𝑂4)2 + 𝐻2𝑆𝑂4 → 𝐶𝑎(𝐻2𝑃𝑂4)2 + 𝐶𝑎𝑆𝑂4
................200 g............133.5 g...........?
You must identify the LR.
1. Balance the equation.
𝐶𝑎3(𝑃𝑂4)2 + 2𝐻2𝑆𝑂4 → 𝐶𝑎(𝐻2𝑃𝑂4)2 + 2𝐶𝑎𝑆𝑂4

2. Convert each reagent to mols. mol = grams/molar mass
mols Ca3(PO4)2 = 200/molar mass Ca3(PO4)2 = ?
mols H2SO4 = 133.5 g/molar mass H2SO4 = ?

3. Convert mols each (as if the other reagent didn't exist) to mols of the product.
mols Ca(H2PO4)2 = mols Ca3(PO4)2 x [1 mol Ca(HPO4)2/1 mol Ca3(PO4)2] or
mols Ca(HPO4)2 = mols H2SO4 x (1 mol Ca(HPO4)2/2 mol H2SO4)

4. In LR problems the one producing the SMALLER number of mols of the product is the correct one to choose.

5. The smaller number of moles x molar mass Ca(H2PO4)2 = grams of Ca(HPO4)2 produced.

Post your work if you get stuck and DON'T use that weird format please.

Answer 2

To find out how much Ca(H2PO4)2 can be produced, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in the reaction and limits the amount of product that can be formed.

1. Calculate the moles of Ca3(PO4)2 and H2SO4 using their given masses and molar masses:
Molar mass of Ca3(PO4)2 = (3 * atomic mass of Ca) + (2 * atomic mass of PO4)
= (3 * 40.08 g/mol) + (2 * (1 * 31.99 g/mol + 4 * 16.00 g/mol))
= 310.18 g/mol

Moles of Ca3(PO4)2 = (mass of Ca3(PO4)2) / (molar mass of Ca3(PO4)2)
= 200.0 g / 310.18 g/mol

Molar mass of H2SO4 = (2 * atomic mass of H) + (1 * atomic mass of S) + (4 * atomic mass of O)
= (2 * 1.01 g/mol) + (1 * 32.07 g/mol) + (4 * 16.00 g/mol)
= 98.09 g/mol

Moles of H2SO4 = (mass of H2SO4) / (molar mass of H2SO4)
= 133.5 g / 98.09 g/mol

2. Calculate the mole ratio of Ca3(PO4)2 and H2SO4 from the balanced equation:
From the reaction equation:
Ca3(PO4)2 + H2SO4 → Ca(H2PO4)2 + CaSO4

The mole ratio between Ca3(PO4)2 and H2SO4 is 1:1.

3. Determine the limiting reactant:
Compare the moles of Ca3(PO4)2 and H2SO4. The reactant with the smaller number of moles is the limiting reactant.

Let's calculate:

Moles of Ca3(PO4)2 = (200.0 g / 310.18 g/mol) = 0.6446 mol
Moles of H2SO4 = (133.5 g / 98.09 g/mol) = 1.3615 mol

Since 0.6446 mol < 1.3615 mol, the Ca3(PO4)2 is the limiting reactant.

4. Calculate the moles of Ca(H2PO4)2 formed:
Since the mole ratio of Ca3(PO4)2 and Ca(H2PO4)2 is 1:1, the moles of Ca(H2PO4)2 formed will be equal to the moles of Ca3(PO4)2.

Moles of Ca(H2PO4)2 = Moles of Ca3(PO4)2 = 0.6446 mol

5. Calculate the mass of Ca(H2PO4)2 formed:
Mass of Ca(H2PO4)2 = Moles of Ca(H2PO4)2 * Molar mass of Ca(H2PO4)2

Molar mass of Ca(H2PO4)2 = (atomic mass of Ca) + (2 * atomic mass of H) + (atomic mass of PO4)
= 40.08 g/mol + (2 * 1.01 g/mol) + (1 * 31.99 g/mol + 4 * 16.00 g/mol)
= 234.05 g/mol

Mass of Ca(H2PO4)2 = 0.6446 mol * 234.05 g/mol

Therefore, the mass of Ca(H2PO4)2 that can be produced is equal to:

Mass of Ca(H2PO4)2 = 151.91 g