Question

K total is 1.11E12

An experiment is created where 0.110 moles of Cu2+ and 0.495 moles of NH3 are combined in water to form 1.00 L of solution. These react according to the process in part 1. Create an ICE table and react the two substances COMPLETELY. After, determine which molecule is in excess, and enter the molarity of this substance below.

Answer 1

mol of Cu+2 = 0.11 mol

mol of NH3 = 0.495 mol

Cu+2(aq) + 4NH3(aq) = Cu(NH3)4+2(aq)

note that ratio is:

1:4

0.11 mol of Cu requires 4*0.11 = 0.44 mol of NH3, which we DO have, so NH3 is in excess

initially

Cu+2(aq) = 0.11

NH3(aq) = 0.495

Cu(NH3)4+2(aq) = 0

Change

Cu+2(aq) = - x

NH3(aq) = -4x

Cu(NH3)4+2(aq) = + x

inequilibrium

Cu+2(aq) = 0.11 - x

NH3(aq) = 0.495 -4x

Cu(NH3)4+2(aq) = 0 + x

we know

Cu+2(aq) = 0.11 - x = 0.11 (they react completely)

NH3(aq) = 0.495 -4x

Cu(NH3)4+2(aq) = 0 + x

0.495-0.440 = 0.055 mol of NH3 are in excess

[NH3] excess = mol/V = 0.055/1 = 0.055 M