Question

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?

Answer 1

Answer:

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