Question

For the reactions shown below, we added 3.75 mL of 0.0450 M Na₂S to a test tube containing one of the two cations (Cu²⁺ or Cd²⁺) and recovered 0.0161 g of precipitate.

Cu(NO₃)₂(aq) + Na₂S(aq) → CuS(s) + 2 NaNO₃(aq)
Cd(NO₃)₂(aq) + Na₂S(aq) → CdS(s) + 2 NaNO₃(aq)

How much precipitate in moles would be recovered theoretically if the ion was Cu²⁺? (Enter an unrounded value. Use at least one more digit than given.)
mol

How much precipitate in moles would be recovered theoretically if the ion was Cd²⁺? (Enter an unrounded value. Use at least one more digit than given.)
mol

How much precipitate in grams would be recovered theoretically if the ion was Cu²⁺?
g

How much precipitate in grams would be recovered theoretically if the ion was Cd²⁺?

g

Based on the precipitate amount recovered, which of the two ions was in the unknown?

Cu²⁺

Cd²⁺

Answer 1

Cu(NO₃)₂(aq) + Na₂S(aq) → CuS(s) + 2 NaNO₃(aq)

no of moles of Na2S = molarity * volume in L

                                 = 0.045*0.00375 = 0.000168moles

1moles of Na2S produce 1 mole of CuS

0.000168 moles of Na2S produce 0.000168 moles of CuS

mass of CuS =no of moles * gram molar mass

                   = 0.000168*95.55 = 0.01605g

Cd(NO₃)₂(aq) + Na₂S(aq) → CdS(s) + 2 NaNO₃(aq)

1 mole of na2S produce 1 moles of CdS

0.000168 moles of Na2S produce 0.000168 moles of CdS

mass of CdS = no of moles * gram molar mass

                      = 0.000168*144.4 = 0.0242g

mass of Cu⁺²   = 0.000168*63.5 = 0.01066g of Cu⁺²

mass of Cd⁺²   = 0.000168*112.4 = 0.0188g of Cd⁺²