How do I figure out what the limiting reagent is for this lab?
Mass of the empty dry beaker =105.52 g
Initial mass of the beaker + CuCl2 (s) = 113.02 g
Initial mass of the two iron nails = 3.82 g
Final mass of the dry beaker and copper = 106.41 g
Final mass of the two iron nails = 3.04 g
1 Answer | Add Yours
From the given data the reaction is: CuCl2 + Fe = FeCl2 + Cu
Mass (in gm, initially taken): CuCl2 (7.5), Fe (3.82)
Moles of reagents, initially taken: CuCl2 (7.5/134.5=0.056), Fe (3.82/55.85=0.068)
Evidently CuCl2 is the limiting reagent here. This is further corroborated from the fact that some portion of the iron nails is recovered unchanged, even after whole of the CuCl2 had reacted. This is the excess reagent. Amount of iron reacted = 3.82-3.04 = 0.78g = 0.78/55.85=0.014 moles. This should be equal to the amount of Cu (metal) formed. Amount of Cu produced actually = 106.41-105.52 = 0.89g = 0.89/63.5 = 0.014 moles. Hence all the data is supporting the concept of CuCl2 as the limiting reagent.
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes