Data: Mass (g) A) Empty 250-mL beaker ll 7 65 B) Solid CuCl2 C) Fe nail before t
ID: 532733 • Letter: D
Question
Explanation / Answer
1. 2Fe + 3CuCI2 >>> 2 FeCI3 + 3Cu+
2. no of moles of solid CuCI2 = 1.28/134.45 g ( mol wt) = 0.0052
Fe (ll)= 2.760/71.844 = 0.03842
solid CuCl2 becomes the limiting reagent. so no of moles limiting reagent= moles of copper produced
theoretical yeild= moles of copper x molar mass of copper
= 0.00952 x 63.546
= 0.604
percent yeild = actual yield / theoretical yeild x 100 %
0.29/0.604 x 100
= 48 %
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