Question

Consider the following balanced chemical equation. 2 K3PO4 (aq) + 3 MgBr2(aq) + Mg3(PO4)2(s) + 6 KBr (aq) A 91.06 mL sample of 0.313 M K3PO4 is mixed with 63.81 mL of 1.972 M MgBr2. After reaction, what is the concentration of the excess reactant (in M)?

Answer 1

volume of K3PO4, V = 91.06 mL

= 9.106*10^-2 L

use:

number of mol in K3PO4,

n = Molarity * Volume

= 0.313*9.106*10^-2

= 2.85*10^-2 mol

volume of MgBr2, V = 63.81 mL

= 6.381*10^-2 L

use:

number of mol in MgBr2,

n = Molarity * Volume

= 1.972*6.381*10^-2

= 0.1258 mol

Balanced chemical equation is:

2 K3PO4 + 3 MgBr2 ---> Mg3(PO4)2 + 6 KBr

2 mol of K3PO4 reacts with 3 mol of MgBr2

for 2.85*10^-2 mol of K3PO4, 4.275*10^-2 mol of MgBr2 is required

But we have 0.1258 mol of MgBr2

so, K3PO4 is limiting reagent

According to balanced equation

mol of MgBr2 reacted = (3/2)* moles of K3PO4

= (3/2)*2.85*10^-2

= 4.275*10^-2 mol

mol of MgBr2 remaining = mol initially present - mol reacted

mol of MgBr2 remaining = 0.1258 - 4.275*10^-2

mol of MgBr2 remaining = 8.308*10^-2 mol

Total volume = 91.06 mL + 63.81 mL = 154.87 mL = 0.15487 L

Use:

[MgBr2] remaining = mol of MgBr2 remaining / Total volume

= (8.308*10^-2 mol) / 0.15487 L

= 0.536 M

Answer: 0.536 M