Force Completion Once started, this test must be completed in one siting Do not

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Force Completion Once started, this test must be completed in one siting Do not leave the test before dicking Save and Submit Remaining Time: 1 hour, 55 minutes, 59 seconds. Question Completion Status: Moving to another question will save this response stion 1 Copy of 5 po Calculate the moles of mercaptan from the limiting reactant in the following synthesis when 8 ml of ethanediol d 1.32 g/ml is mixed with 15 gms of mercury d-5.43 g/ ethanediol m mercury ion mercaptan 001 moles O 0.085 moles 0 0.075 moles O 0.056

Explanation / Answer

mol of ethanediol = masS/MW =(D*V)/MW = (1.32*8)/(62) = 0.170322 mol of ethanediol

mol of Hg = mass/MW = (D*V)/MW = 15/200.59 = 0.07477 mol of Hg

ratio is 1:1 so

Hg is limiting

0.07477 mol of Hg will form 0.07477 mol of Mercaptan

from the list

choose 0.075 mol of product

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