H3MO3 was titrated with 6.67 mL of 0.100 M MnO4^-

H3MO3 was titrated with 6.67 mL of 0.100 M MnO4^-

 

 

On heating a 0.200 g sample of the semimetal in the air, the corresponding oxide M2O3 was obtained. When the oxide was dissolved in aqueous acid, H3MO3 was formed. When this was titrated with KMnO4, 6.67 mL of 0.100 M MnO4^- was required to reach an equivalence point. The unbalanced equation is

H3MO3(aq) + MnO4^-(aq) → H3MO4(aq) + Mn^2+(aq) (in acid)

a. Balance the equation.

b. How many moles of semimetal M were in the initial 0.200 g sample?

c. What is the identity of the semimetal M?

 

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a. Balance the equation.

5H3MO3 + 2MnO4^- + 6H^+ → 5H3MO4 + 2Mn^2+ + 3H2O

5H3MO3(aq) + 2MnO4^-(aq) + 6H^+(aq) → 5H3MO4(aq) + 2Mn^2+(aq) + 3H2O(l)

( 설명 https://ywpop.tistory.com/11693 )

 

 

 

b.

6.67 mL of 0.100 M MnO4^-

(0.100 mol/L) × 0.00667 L = 0.000667 mol MnO4^-

 

 

H3MO3 : MnO4^- = 5 : 2 계수비(= 몰수비) 이므로,

H3MO3 : MnO4^- = 5 : 2 = ? mol : 0.000667 mol

? = 5 × 0.000667 / 2 = 0.00167 mol H3MO3

 

 

H3MO3 : M = 1 : 1 몰수비 이므로,

semimetal M의 몰수 = 0.00167 mol

 

 

 

c

0.200 g / 0.00167 mol = 119.76 g/mol

Sb의 몰질량 = 121.76 g/mol

답은 Sb.

 

 

 

[ 관련 예제 https://ywpop.tistory.com/11694 ]