Consider the following balanced chemical equation: 2K3PO4 (aq) + 3MgBr2 (aq) ----> Mg3(PO4)2 (s) + 6KBr (aq)

Question

A 97.17 mL sample of 1.4 M K₃PO₄ is mixed with 45.24 mL of 0.615 M MgBr₂. After reaction, what is the concentration of the excess reactant (in M)?

Answer 1

2K₃PO₄(aq) + 3MgBr₂(aq) ===> Mg₃(PO₄)₂(s) + 6KBr(aq) ... balanced equation

Finding limiting reactant:

moles K₃PO₄ present = 97.17 ml x 1 L/1000 ml x 1.4 mol/L = 0.1360 moles

moles MgBr₂ present = 45.25 ml x 1 L/1000 ml x 0.615 mol/L = 0.02783 moles

Easy way to find limiting reactant is to divide moles by coefficient in balanced equation and see which is less:

For K₃PO₄ we have 0.1360 / 2 = 0.068

For MgBr₂ we have 0.02783 / 3 = 0.0093 <== LIMITING

So, now we know that MgBr₂ is limiting, meaning that K₃PO₄ is in excess. The question asks for the concentration of K₃PO₄ after the reaction.

moles K₃PO₄ used up during reaction: 0.02783 mol MgBr₂ x 2 mol K₃PO₄/3 mol MgBr₂ = 0.018555 moles

of K₃PO₄ used up.

Moles K₃PO₄ left over = 0.1360 moles - 0.01856 moles = 0.1174 moles K₃PO₄ left over.

What is the final volume? 97.17 ml + 45.24 ml = 142.41 mls = 0.14241 L

Final [K₃PO₄] = 0.1174 moles K₃PO₄ / 0.1424 L = 0.82 M (to 2 significant figures)