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

70.2K

Verified Solution

Question

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) â†’ CuS(s) + 2NaNO₃(aq)
Cd(NO₃)₂(aq) +Na₂S(aq) â†’ 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

Get Answers to Unlimited Questions

Join us to gain access to millions of questions and expert answers. Enjoy exclusive benefits tailored just for you!

Membership Benefits:

Unlimited Question Access with detailed Answers
Zin AI - 3 Million Words
10 Dall-E 3 Images
20 Plot Generations
Conversation with Dialogue Memory
No Ads, Ever!
Access to Our Best AI Platform: Flex AI - Your personal assistant for all your inquiries!

Become a Member