Question

Solve the following using the reactions:

5C6H12O6 + 24NO3^- + 24H^+ --> 30CO2 + 12N2 + 42H2O

C6H12O6 = 4CrO7^2- + 32H^+ --> 8Cr^3+ + 6CO2 + 22H2O

How much N2 will be produced (L) when 6 liters of a C6H12O6 solution is degraded to CO2 and H2O? The concentration of C6H12O6 solution is 200 mg/L

What concentration (mg/L) of Cr2O72- needs to be added to degrade 200 mg/L to C6H12O6 to CO2?

Answer 1

1)

volume of C6H12O6 = 6 L

concentration = 200 mg / L = 0.200 g /L

                      = 0.200 / 180

                     = 1.11 x 10^-3 M

moles of C6H12O6 = 1.11 x 10^-3 x 6

                                 = 6.66 x 10^-3 mol

5 mol C6H12O6 ----------> 12 mol N2

6.66 x 10^-3 mol --------> ??

moles of N2 = 6.66 x 10^-3 x 12 / 5 = 0.01598 mol

mass of N2 produced = 0.01598 x 28

amount of N2 produced = 0.448 g

2)

concentration = 200 mg / L = 0.200 g /L

                      = 0.200 / 180

                     = 1.11 x 10^-3 M

moles of C6H12O6 = 1.11 x 10^-3 x 6

                                 = 6.66 x 10^-3 mol

1 mol C6H12O6 --------------> 4 mol C2rO7

6.66 x 10^-3 mol ----------> ??

moles of Cr2O7 = 6.66 x 10^-3 x 4 = 0.02664 mol

concentration = 4.44 x 10^-3 mol / L

                        = 4.44 x 10^-3 x 216

                         = 0.959 g / L

cocnentration of Cr2O72- = 959 mg / L