Glucose (C6H12O6) and ammonia (NH3) are anaerobically converted(fermented) by ye

Question

Glucose (C6H12O6) and ammonia (NH3) are anaerobically converted(fermented) by yeast (CH1.74N0.2O0.45) to ethanol (C2H5OH),glycerol (C3H8O3), carbon dioxide (CO2), and water (H2O) accordingto the empirical stoichiometric equation
C6H12O6 + a(NH3) b(CH1.74N0.2O0.45) + c(C2H5OH) + d(C3H8O3)+ e(CO2) + f(H2O)
Experimental data indicates that 0.12 moles of glycerol are formedfor each mole of ethanol produced and that 0.08 moles of water areformed for each mole of glycerol.
a) Derive a set of linear equations that can be used to determinethe values of the stoichiometric coefficients (a, b, c, d, e, &f) in the relation above.
b) Find the corresponding values for a, b, c, d, e, & f.

Explanation / Answer

Question Details: Glucose (C6H12O6) and ammonia(NH3) are anaerobically converted (fermented) by yeast (CH1.74N0.2O0.45) to ethanol (C2H5OH),glycerol (C3H8O3), carbon dioxide (CO2), and water (H2O) according to the empiricalstoichiometric equation C6H12O6 + a(NH3) b(CH1.74N0.2O0.45) + c(C2H5OH) + d(C3H8O3)+ e(CO2) + f(H2O) Experimental data indicates that 0.12 moles of glycerol are formedfor each mole of ethanol produced and that 0.08 moles of water are formed for eachmole of glycerol. a) Derive a set of linear equations that can be used to determinethe values of the stoichiometric coefficients (a, b, c, d, e, & f) in therelation above. C6H12O6 + a(NH3) b(CH1.74N0.2O0.45) + c(C2H5OH) + d(C3H8O3)+ e(CO2) + f(H2O) CARBON BALANCE 6=B+2C+3D+E............1 HYDROGEN BALANCE 12=-3A+1.74B+5C+8D+2F..................................2 OXYGEN BALANCE 6=0.45B+C+3D+2E+F...................................................3 NITROGEN BALANCE 0=-A+0.2B..........................................4 Experimental data indicates that 0.12 moles of glycerol are formedfor each mole of ethanol produced ....THAT IS D=0.12....................5 and that 0.08 moles of water are formed for each mole ofglycerol....THAT IS F/D=0.08...OR F=0.08*0.12=0.0096..............................6 HENCE EQNS. 1 TO 4 BECOME 6=B+2C+3D+E................5.64=B+2C+E.......................................7 12=-3A+1.74B+5C+8D+2F..............12-0.9600-0.0192=11.0208=-3A+1.74B+5C...................8 6=0.45B+C+3D+2E+F..................5.6304=0.45B+C+2E..........................9 0=-A+0.2B..........................................4 b) Find the corresponding values for a, b, c, d, e, & f. M*X=K M= X= K= 0 1 2 1 A 5.64 -3 1.74 5 0 * B = 11.0208 0 0.45 1 2 C 5.6304 -1 0.2 0 0 E 0 X=M INVERSE * K M INVERSE= 0.461893764 -0.138568129 -0.230946882 -0.584295612 2.309468822 -0.692840647 -1.154734411 2.07852194 -0.526558891 0.357967667 0.263279446 -1.073903002 -0.256351039 -0.023094688 0.62817552 0.069284065 SO WE HAVE.........X= A= -0.222374134 B= -1.11187067 C= 2.457666513 E= 1.836537644 D=0.12 F=0.0096

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