Question

For the reactions shown below, we added 3.75 mL of 0.0420 M Na₂S to a test tube containing one of the two cations (Cu²⁺ or Cd²⁺) and recovered 0.0228 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⁺?


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


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

Answer 1

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

a.From the equation given above

1 mol Na2S will produce 1 mol CuS

Now we have to know the moles of Na2S

We know

Mole = Volume in liter x Molarity of the solution

         

Volume of Na2S = 3.75mL = 3.75/1000 L

Molarity of Na2S = 0.0420M

Mole of Na2S = 3.75/1000 x 0.0420 = 0.0001575 mol Na2S

0.0001575 mol Na2S will produce 0.001575 mol CuS

b. this is same as CuS

boz here also 1 mole of Na2S will produced 1 mole of CdS

Therefore

0.0001575 mol Na2S will produce 0.001575 mol CdS

c. in this the Molar mass CuS = 63.55 (Cu)+32(S) = 95.55g/mol

mass of CuS in grams = mole of CuS x Molar mass of CuS

                                     = 0.0001575 mol x 95.55 g/mol

                                     = 0.0150g CuS precipitated. Theoretical yield.

d. in this the Molar mass CdS = 112.4 (Cd)+32(S) = 144.4g/mol

mass of CdS in grams = mole of CdS x Molar mass of CdS

                                     = 0.0001575 mol x 1444 g/mol

                                     = 0.0227g CdS precipitated. Theoretical yield.

e. the amount of CdS is very close to the actual precipitate mass . The solution is more likely to contain Cd(NO3)2 than Cu(NO3)2.