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
