48.0 mL of 1.70 M CuCl2 (aq) and 57.0 mL of 0.800 M (NH4) 2S (aq) are mixed together to give CuS (s) as a precipitate. The other product of the reaction is aqueous ammonium chloride. What is the concentration of the Cu (II) ion after the complete reaction?

The concentration of the Cu (II) ion is 0.777M

Explanation:

Step 1: Data given

Volume of 1.70 M CuCl2 = 48.0 mL = 0.0480 L

Volume of 0.800 M (NH4) 2S = 57.0 mL = 0.0570 L

Step 2: The balanced equation

CuCl2 (aq) + (NH4) 2S (aq) → 2 NH4Cl (aq) + CuS (s)

Step 3: Calculate moles CuCl2

moles CuCl2 = 0.0480 L * 1.70 M=0.0816 moles

Step 4: Calculate moles (NH4) 2S

moles (NH4) 2S = 0.0570 L * 0.800 M = 0.0456 moles

Step 5: Calculate the limiting reactant

The ratio between CuCl2 and (NH4) 2S is 1 : 1 so (NH4) 2S is the limiting reactant. IT will completely be consumed (0.0456 moles).

CuCl2 is in excess. There will remain 0.0816 - 0.0456 = 0.0360 moles

Step 6: Calculate moles of CuS

For 1 mol CuCl2 we need 1 mol (NH4) 2S to produce 2 moles of NH4Cl and 1 mol CuS

For 0.0456 moles we'll produce 0.0456 moles CuS

Step 7: Calculate moles of Cu (II) ion

There remains 0.0360 moles CuCl2.

There will be 0.0456 moles CuS produced

Total moles Cu^2 + = 0.0816 moles

Step 8: Calculate concentration of Cu (II) ion

Concentration = moles / volume

Concentration = 0.0816 moles / (0.048+0.057)

Concentration = 0.777 M

The concentration of the Cu (II) ion is 0.777M