Answer:
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
