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.
..............πΆπ3(ππ4)2 + π»2ππ4 β πΆπ(π»2ππ4)2 + πΆπππ4
................200 g............133.5 g...........?
You must identify the LR.
1. Balance the equation.
πΆπ3(ππ4)2 + 2π»2ππ4 β πΆπ(π»2ππ4)2 + 2πΆπππ4
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.
Superphosphate, or sometimes called triple phosphate, is a commonly used water-soluble
fertilizer. It is produced by the reaction:
πΆπ3
(ππ4
)2 + π»2ππ4 β πΆπ(π»2ππ4
)2 + πΆπππ4
If we mix 200.0 g of πΆπ3
(ππ4
)2 with 133.5 g of π»2ππ4, how much πΆπ(π»2ππ4
)2 (in grams)
can be produced?
1 answer