Question

Balance the following reactions using a half-reaction method: (H- or H20 might need to be added) C6H1206 +NO N2 +CO2 C6H1206+ Cr207 Cr CO2+ H20 Solve the following using the reactions you balanced above (Make an assumption if it is needed) a. How much N2 will be produced (L) when 6 liters of a CsH120 solution is degraded to CO2 and H20? The concentration of CoH120s solution is 200 mg/L b. What concentration (mg/L) of Cr2072- needs to be added to degrade 200 mg/L to C6H1206 to CO2?

I balanced them and got the following:

5c6h12o6+24No3- +24H+----->30Co2+12N2+42h2o

and

c6h12o6+4Cr2o7(2-)+32H+---6Co2+8Cr(3+)+22h2o

the 2- and 3+ are superscripts. thank you

Answer 1

Your reaction is correctly balanced

Oxidation: C_6H_12O_6 + 6 H_2O rightarrow 6 CO_2 + 24 H^+ + 24 c times 5 2 NO^-3_3 + 12H^+ rightarrow N_2 + 6 H_2O times 12/5 C_6H_12O_6 + 24 NO^-_3 + 24 H^+ rightarrow 3 OCO_2 + 12 N_2 + 42 H_2O 2nd reaction C_6H_12O_6 + 4 cr_2O^2-_7 + 32 H^+ rightarrow 6 CO_2 + 8 cr^3 + + 22 H_2O (a) total mass of C_6H_12O_6 = 200 times 6 = 1200 mg = 1.2 g moles = 1.2/180 = 6.67 times 10^-3 let moles of M_2 = x x/12 = 6.67 times 10^-3/5 = > x = 16 times 10^-3 assuming N_2 at STP, volume = 16 times 10^-3 times 22.4 = 0.358: L 200 mg/L C_6H_12O_6 in term of molarity, 200 times 10^-3/180 mol/L for 1 mole C_6H_12O_6 we need, 4 mole Cr_2O^2-_7 so cr_2O^2-_7 need = 4 times 200 times/180 216 mg/L = 960 mg/L