Question

what is the maximum mass of Copper (II) sulfide which can be formed when 38.0 mL of 0.500M CuCl2 are mixed with 42.0mL of 0.600M (NH4)2S? Aqeuous ammonium chloride is the other product.

Answer 1

CuCl2(aq) + (NH4)2S(aq) --> CuS(s) + 2NH4Cl(aq)

First calculate the mole sof both regent as follows:

Number of mole = molarity * volume in L

CuCl2: (0.500 M)(38/1000 L) = 0.019 moles
(NH4)2S: (0.600 M)(42/1000 L) = 0.0252 moles

Now determine the limiting reagent;

CuCl2 is limiting reagent , The limiting agent has due to following properties:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

then calculate the mole OF Cus as follows:

(0.019 moles)(1 mol CuS/ 1 mol CuCl2) = 0.019 moles CuS

amount of CuS will be calculated as follows:

Amount in g = molar mass * number of moles

= 95.611 g/mole *0.019 moles CuS

=1.82 g CuS
molar mass of CuS = 95.611 g/mole