1.67×10^(-3) M C6H12O6 oxidized to CO2 and H2O

1.67×10^(-3) M C6H12O6 oxidized to CO2 and H2O

 

 

Consider a 1.67×10^(-3) M glucose solution (C6H12O6)

that is completely oxidized to CO2 and H2O.

Find the amount of oxygen required to complete the reaction.

 

Find the amount of oxygen required

in mg/L to complete the reaction.

 

 

 

1.67×10^(-3) M C6H12O6 용액

= 1.67×10^(-3) mol C6H12O6 / 1 L 용액

---> 용액 1 L당 1.67×10^(-3) mol C6H12O6

 

 

 

C6H12O6 + 6O2 → 6CO2 + 6H2O

( 참고 https://ywpop.tistory.com/9594 )

 

C6H12O6 : O2 = 1 : 6 = 1.67×10^(-3) mol : ? mol

 

? = 6 × (1.67×10^(-3)) = 0.01002 mol O2

---> 용액 1 L당 0.01002 mol O2

 

 

 

0.01002 mol × (32.00 g/mol) = 0.32064 g O2

 

 

 

0.32064 g × (1000 mg / 1 g) = 320.64 mg O2

---> 용액 1 L당 320.64 mg O2

 

 

 

답: 321 mg/L