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