Determine the Mass of CuS formed when 100. mL of 0.100 M H2S and 50.0 mL of 0.150 M Cu(C2H3O2) are reacted. What is the concentration of?
C2H3O2- in the resulting solution? I got 0.717 g CuS and 0.100 M C2H3O2-
- hcbiochemLv 71 month ago
The second compound should be Cu(C2H3O2)2.
For C2H3O2-, its initial concentration is 0.300 M. So:0.300 M (50 mL) = M2(150 mL)M2 = 0.100 M
Moles Cu2+ = 0.0500 L X 0.150 mol/L = 7.5X10^-3 mol
Moles S2- = 0.100 L X 0.100 mol/L = 0.0100 mol
Cu2+ is the limiting reactant and you can form only 7.5X10^-3 mol CuS
Mass CuS = 7.5X10^-3 mol X 95.61 g/mol = 0.717 g CuS
You've got this.