0.3000 g Sn 8.08 mL 0.0500 M KMnO4

0.3000 g Sn 8.08 mL 0.0500 M KMnO4

 

 

A sample of a tin ore weighing 0.3000 g

was dissolved in an acid solution

and all the tin in the sample was changed to tin(II).

The solution was titrated with 8.08 mL of 0.0500 M KMnO4 solution,

which was oxidized the tin(II) to tin(IV).

What was the percentage by weight of tin in the sample?

 

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3Sn^2+(aq) + 2MnO4^-(aq) + 8H^+(aq)

→ 3Sn^4+(aq) + 2MnO2(s) + 4H2O

 

 

 

(0.0500 mol/L) (8.08/1000 L) = 0.000404 mol KMnO4

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

Sn^2+ : MnO4^- = 3 : 2 = ? mol : 0.000404 mol

? = 3 × 0.000404 / 2 = 0.000606 mol Sn^2+

 

 

 

Sn의 몰질량 = 118.71 g/mol

0.000606 mol × (118.71 g/mol) = 0.0719 g Sn

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

(0.0719 / 0.3000) × 100 = 23.97%

 

 

 

답: 24.0%

 

 

 

 

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