20.0 g H2O2 0.145 M KMnO4 46.9 mL H2O2 mass percent

20.0 g H2O2 0.145 M KMnO4 46.9 mL H2O2 mass percent

 

 

20.0 g H2O2 용액 적정에

0.145 M KMnO4 용액 46.9 mL 필요하다면

H2O2의 질량 백분율은 얼마인가?

 

 

A solution of hydrogen peroxide, H2O2, is titrated

with a solution of potassium permanganate, KMnO4.

The reaction is :

5H2O2 + 2KMnO4 + 3H2SO4 → 5O2 + 2MnSO4 + K2SO4 + 8H2O

It requires 46.9 mL of 0.145 M KMnO4

to titrate 20.0 g of the solution of hydrogen peroxide.

What is the mass percentage of H2O2 in the solution?

 

 

A solution of H2O2 is titrated against a solution of KMnO4.

The reaction is :

2MnO4^- + 5H2O2 + 6H^+ → 2Mn^2+ + 5O2 + 8H2O

If it requires 46.9 mL of 0.145 M KMnO4 to oxidise 20 g of H2O2,

the mass percentage of H2O2 in this solution is _____.

 

---------------------------------------------------

 

적정에 필요한 KMnO4의 몰수를 계산하면,

(0.145 mol/L) (46.9/1000 L) = 0.0068005 mol KMnO4

 

 

 

분자 반응식

5H2O2 + 2KMnO4 + 3H2SO4 → 5O2 + 2MnSO4 + K2SO4 + 8H2O

( 참고 https://ywpop.tistory.com/8703 )

 

알짜 이온 반응식

5H2O2 + 2MnO4^- + 6H^+ → 5O2 + 2Mn^2+ + 8H2O

 

 

 

H2O2 : KMnO4 = 5 : 2 계수비(= 몰수비) 이므로,

0.0068005 mol KMnO4와 반응하는 H2O2의 몰수를 계산하면,

H2O2 : KMnO4 = 5 : 2 = ? mol : 0.0068005 mol

 

? = 5 × 0.0068005 / 2 = 0.01700125 mol H2O2

 

 

 

H2O2의 몰질량 = 34.0147 g/mol 이므로,

0.01700125 mol H2O2의 질량을 계산하면,

0.01700125 mol × (34.0147 g/mol) = 0.5783 g H2O2

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

H2O2의 질량 백분율을 계산하면,

(0.5783 g / 20.0 g) × 100 = 2.89%

 

 

 

답: 2.89%