After collecting the entire class data, you should be able to construct your dat

Question

After collecting the entire class data, you should be able to construct your data table and graph your data. You must have all the columns provided in the sample data table in Part 2 Use the following table to guide you,mi Trial m(AI), g mL0.250 m(Cu), g n(Al) n(Cu) ratio drops 6M | %yield ? M CuSO n(Cu)/n(AI) HCI 25 0 36 0.0057 12,05 54 20 230 00,4 00040,006 1, 4318 3 04 35 0, 5t 0.0052 8 1.5385 Ib 40 30005 0.03544076 Using Excel and the class data you must now graph the following: a) moles of Cu vs. moles of Al o nf Alus moles of aluminum

Explanation / Answer

For trial 1 :
2 Al + 3 CuSO4 - -----> 3 Cu (s) + Al2(SO4)3
Al is the limiting reactant so,
2 mol of Al produces 3 mol of Cu
0.0037 mol of Al produces ? Cu
just cross multiply to get the answer
Theoretical mol of Cu = (0.0037*3)/2 = 5.55*10-3 mol
Theoretical yield of Cu in gm = 5.55*10-3mol *63.5463 g/mol = 0.3526 gm
experimental yield of Cu = 0.36 gm
here exp yield > theoretical yield so formula for % yield will be,
% yield = (theoretical / exp )*100 = (0.3526 / 0.36)*100 = 97.94 %

For trial 2 :
2 mol of Al produces 3 mol of Cu
0.0044 mol of Al produces ? Cu
Theoretical mol of Cu = (0.0044*3)/2 = 6.6*10-3 mol
Theoretical yield of Cu in gm = 6.6*10-3mol*63.5463 gm/mol = 0.4194 gm
experimental yield = 0.4 gm
% yield = (0.4 / 0.4194) *100 = 95.37 %

For trial 3 :
2 mol of Al produces 3 mol of Cu
0.0052 mol of Al produces ? Cu
Theoretical mol of Cu = (0.0052*3)/2 = 7.8*10-3mol
Theoretical yield of Cu in gm = 7.8*10-3mol * 63.5463gm/mol = 0.4956 gm
experimental yield = 0.51 gm
% yield = ( 0.4956 / 0.51) *100 = 97.17 %

For trial 4 :
2 mol of Al produces 3 mol of Cu
0.0059 mol of Al produces ? Cu
Theoretical mol of Cu = (0.0059*3)/2 = 8.85*10-3 mol
Theoretical yield of Cu in gm = 8.85*10-3 mol *63.5463gm/mol = 0.5623 gm
experimental yield = 0.543 gm
% yield = (0.543 / 0.5623) *100 = 96.56 %
Similarly you can calculate for trials 5 & 6 as mentioned in above 4 trials.

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