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

Question

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.