Question

Consider the following reaction: 3C2H5OH + 2 Na2Cr2O7 + 8H2SO4 à 3CH3COOH + 2Cr2(SO4)3 + 2Na2SO4 + 11 H2O

a) If a reactor is fed with a mixture having the following composition (in mol%): 20% de C2H5OH, 20% de Na2Cr2O7, and the balance H2SO4, which is the limiting reactant?

b) If 230 kg/h of C2H5OH are fed to the reactor, what is the amount (in kg/h) of the other two reactants that need to be fed to have a stochimetric feed?

Answer 1

Basis - 100 moles of mixture

Part a

Moles of C2H5OH = 20

Moles of Na2Cr2O7 = 20

Moles of H2SO4 = 60

The balanced reaction

3C2H5OH + 2 Na2Cr2O7 + 8H2SO4 - - - - - > 3CH3COOH + 2Cr2(SO4)3 + 2Na2SO4 + 11 H2O

From the stoichiometry of the reaction

3 moles of C2H5OH require = 2 mol of Na2Cr2O7 = 8 mol of H2SO4

20 moles of C2H5OH require =( 2*20/3=) 13.33 mol of Na2Cr2O7 = (8*20/3=) 53.33 mol of H2SO4

All 20 moles of C2H5OH consumed in the reaction

H2SO4 consumed (53.33 mol) less than we have (60mol)

Na2Cr2O7 consumed (13.33 mol) less than we have (20mol)

Limiting reactant = C2H5OH

Part b

From the stoichiometry of the reaction

3 moles of C2H5OH require = 2 mol of Na2Cr2O7 = 8 mol of H2SO4

(3*46=) 138 g C2H5OH require =( 2*261.97=) 523.94 g of Na2Cr2O7 = (8*98=) 784 g of H2SO4

230 g C2H5OH require = 523.94*230/138 = 873.23 g of Na2Cr2O7

230 kg/h C2H5OH require = 873.23 kg/h of Na2Cr2O7

230 g C2H5OH require = 784*230/138 = 1306.67 g of H2SO4

230 kg/h C2H5OH require = 1306.67 kg/h of H2SO4

Amount of H2SO4 required = 1306.67 kg/h

Amount of Na2Cr2O7 required = 873.23 kg/h