How many grams of calcium phosphate are theoretically produced if we start with 3.40 moles of ca(no3)2 and 2.40 moles of li3po4? reaction: 3ca(no3)2 + 2li3po4 → 6lino3 + ca3(po4)2

527.3 g of Ca₃(PO₄)₂ would be produced in the reaction

Explanation:

Let's analyse the reactions and the amount of each reactant:

3Ca(NO₃)₂ + 2Li₃PO₄  →  6LiNO₃  + Ca₃(PO₄)₂

3.4 moles      2.40 mol

3 moles of calcium nitrate react with 2 moles of lithium phosphate.

So 3.4 moles of calcium nitrate, would need (3.4  .2)/3 = 2.26 moles to react

We have 2.40 moles, so the lithium phosphate is the reactant in excess.

Then, the limiting is the  Ca(NO₃)₂. Let's verify it.

2 moles of lithium phosphate need 3 moles of Ca(NO₃)₂ to react

So, 2.4 moles of lithium phosphate would need (2.4 .3)/ 2 = 3.6 moles to react. We have only 3.40 moles, that's why the limiting is the  Ca(NO₃)₂.

Now we can solve the problem.

2 moles of Calcium nitrate produce 1 mol of calcium phosphate

So, 3.4 moles of calcium nitrate, would produce (3.4  .1)/2 = 1.7 moles if Ca₃(PO₄)₂

Let's convert the moles to mass (mol . molar mass)

1.7 mol  . 310.18 g/mol = 527.3 g