0.5000 g VO^2+ 26.45 mL 0.02250 M KMnO4 V mass percent

0.5000 g VO^2+ 26.45 mL 0.02250 M KMnO4 V mass percent

 

 

The vanadium (V) ion in a 0.5000 g sample of ore

is converted to VO^2+ ions.

The amount of VO^2+ in solution can be determined

by reaction with an acid solution of KMnO4.

The balanced equation for the reaction is

5VO^2+(aq) + MnO4^-(aq) + 11H2O(l)

→ Mn^2+(aq) + 5V(OH4)^+(aq) + 2H^+(aq)

 

What is the mass percent of vanadium in the ore

if 26.45 mL of 0.02250 M permanganate solution

is required for complete reaction?

 

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반응한 KMnO4의 몰수

= (0.02250 mol/L) (26.45/1000 L) = 0.0005951 mol KMnO4

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

5VO^2+(aq) + MnO4^-(aq) + 11H2O(l)

→ Mn^2+(aq) + 5V(OH4)^+(aq) + 2H^+(aq)

 

 

 

VO^2+ : MnO4^- = 5 : 1 계수비(= 몰수비) 이므로,

0.0005951 mol KMnO4와 반응한 VO^2+의 몰수를 계산하면,

VO^2+ : MnO4^- = 5 : 1 = ? mol : 0.0005951 mol

 

? = 5 × 0.0005951 = 0.0029755 mol VO^2+

= 0.0029755 mol V

 

 

 

V의 몰질량 = 50.94 g/mol 이므로,

0.0029755 mol × (50.94 g/mol) = 0.15157 g V

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

바나듐의 질량 백분율을 계산하면,

(0.15157 g / 0.5000 g) × 100 = 30.31%

( 참고 https://ywpop.tistory.com/2656 )

 

 

 

답: 30.31%

 

 

 

 

[키워드] redox VO^2+ + MnO4^-, redox MnO4^- + VO^2+