H3MO3 was titrated with 10.7 mL of 0.100 M MnO4^-

H3MO3 was titrated with 10.7 mL of 0.100 M MnO4^-

 

 

공기 중에서 어떤 반금속 샘플 0.200 g을 가열하면, M2O3에 해당하는 산화물이 얻어진다. 이 산화물을 산성 수용액에 용해시켜 KMnO4로 적정할 때, 반응을 완결시키는데 10.7 mL의 0.100 M MnO4^-가 소요되었다. 비균형 방정식은 다음과 같다.

 

 

On heating a 0.200 g sample of a certain semimetal in air, the corresponding oxide M2O3 was obtained. When the oxide was dissolved in aqueous acid and titrated with KMnO4, 10.7 mL of 0.100 M MnO4^- was required for complete reaction. The unbalanced equation is

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

(a) Balance the equation.

(b) How many moles of oxide were formed, and how many moles of semimetal 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)

반응한 MnO4^-의 몰수를 계산하면,

10.7 mL of 0.100 M MnO4^-

(0.100 mol/L) × 0.0107 L = 0.00107 mol MnO4^-

 

 

0.00107 mol MnO4^-와 반응한 H3MO3의 몰수를 계산하면,

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

H3MO3 : MnO4^- = 5 : 2 = ? mol : 0.00107 mol

? = 5 × 0.00107 / 2 = 0.00268 mol H3MO3

 

 

M2O3 + 3H2O → 2H3MO3

M2O3 : H3MO3 = 1 : 2 = ? mol : 0.00268 mol

? = 0.00268 / 2 = 0.00134 mol M2O3

 

 

H3MO3 : M = 1 : 1 몰수비 이므로,

semimetal M의 몰수 = 0.00268 mol

 

 

 

(c)

0.200 g / 0.00268 mol = 74.6 g/mol

( 계산 설명 https://ywpop.tistory.com/7738 )

 

As의 몰질량 = 74.92 g/mol 이므로,

semimetal은 As.