Question

2 C2H5OH + 2 Na2Cr2O7 + 16HCl → 3 CH3COOH + 4 CrCl3 + 4 NaCl + 11 H2O the yield (called a theoretical yield) of CrCl3 we can get from the reaction of 14.0 mol of C2H5OH, 10.0 mol of Na2Cr2O7 and 50.0 mol of HCl would be

Answer 1

In the given equation, c2H5OH and Na2Cr2O7 reacts in equimolar quantities

Ratio of the reactants reacting=2 mol C2H5OH: 2 mol Na2Cr2O7:16 mol HCl

                                                       =1:1:8

1 mol of c2H5OH and Na2Cr2O7 reacts with 8 mol HCl

So 10 mol of each would react with 10*8=80 mol HCl,but HCl is in less amount so is the limiting reagent for the reaction.

As 16 mol HCl reacts to give 4 moles CrCl3

So 50 mol HCl react (with 1/8*50=6.25 moles of c2H5OH and Na2Cr2O7) to give 4/16*50 mol =12.5 mol CrCl3=12.5mol*158g/mol=1975g(theoretical yield)