Question

1. Consider the following reaction: 1.00 g 20.0 mL 6.00 M 14.0 g a) Which substance is the limiting reactant? b) What is the theoretical yield of the complex formed in the reaction? c) If 10.0 g of the complex was obtained from the reaction, what is the percent yield? 2. Draw the structures for the Al(C204)3 ion and the Cu(C204) ion

Answer 1

Al + 3koH + 3H_2c_2 o4.2H_20 rightarrow K_3 Al (c_H_4)_3 3H_2o + 6H_2o + 3/2 H_2 1.0gm 20ml 14.0g mw = 423.28 Now 26.98 40 136 (a) Here limiting reagent Al, because it react with other reagent given one mole of complex: so we first calculate Al moles: - (b) moles of Al = gm wt of Aluminium/molecular weight of aluminium = 1.0g/26.98 = 0.03706 g moles of Aluminium same moles of final complex K_3 Al (c_2 H_4)_3 3H_o generated so moles of complex = K_2 Al(c_2 H_4)_3 3H_2o = 0.0706 g moles from above information we can calculate, theoretical yield of complex = moles of complex times molecular weight of complex = 15.68gm \r\n Theoretical yield of complex = 15.68 (c) % yield = practical yield/Theoretical yield times 100 = 10g/15.68 times 10 % yield = 63.77% (2) structure of complex- Al(c_2o4)^3_3- ion cu(c_204)^2-_2 ion