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For the reactions shown below, we added 3.75 mL of 0.0450 M Na2S to a test...

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Chemistry

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

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

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

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

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

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

g

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

Cu2+

Cd2+

Answer & Explanation Solved by verified expert
3.8 Ratings (525 Votes)

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
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