How many moles of Ca3 (PO4) 2 can be formed when 1.9 moles of CaCO3 react with 2.7 moles of FePO4?

Answer:-

0.633 moles

Explanation:-

The first step to solving this problem is to write the balanced chemical equation.

The balanced chemical equation for this reaction is

3 CaCO3 + 2 FePO4 = Ca3 (PO4) 2 + Fe2 (CO3) 3

From the balanced chemical equation, we can see

3 moles of CaCO3 react with 2 moles of FePO4.

In the question we are told there are 1.9 moles of CaCO3.

So 1.9 moles of CaCO3 react with (2 moles x 1.9 moles) / 3 moles = 1.267 moles of FePO4.

The question tells us there are 2.7 moles of FePO4.

So there is excess FePO4.

Thus our limiting reagent is CaCO3. It will determine how much Ca3 (PO4) 2 is formed.

From the balanced chemical equation, we can see

3 moles of CaCO3 gives 1 mole of Ca3 (PO4) 2

1.9 moles of CaCO3 gives (1 moles x 1.9 moles) / 3 moles = 0.633 moles of Ca3 (PO4) 2.

0.633 moles of Ca3 (PO4) 2 can be formed when 1.9 moles of CaCO3 react with 2.7 moles of FePO4