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  3. for the reactions shown below, we added 3.75 ml of

For the reactions shown below, we added 3.75 mL of 0.0420 M Na2S to a test...

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Chemistry

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


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


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

Answer & Explanation Solved by verified expert
4.1 Ratings (744 Votes)
CuNO32aqNa2SaqCuSs2NaNO3aq CdNO32aq Na2Saq CdSs 2 NaNO3aq aFrom 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    See Answer
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