Fe^2+ 0.2792 g Fe(lll) 0.0194 M K2Cr2O7 23.30 mL mass

Fe^2+ 0.2792 g Fe(lll) 0.0194 M K2Cr2O7 23.30 mL mass

 

 

Fe^2+ 이온만을 포함하는 철광석 0.2792 g을 측정하여

묽은 산에 녹인 후, 모든 Fe(lll) 이온으로 산화시키는데

0.0194 M K2Cr2O7 용액 23.30 mL가 소모되었다.

광석에 포함된 철의 질량 백분율을 구하시오.

Cr2O7^2- + 6Fe^2+ + 14H^+ → 2Cr^3+ + 6Fe^3+ + 7H2O

 

 

A sample of iron ore (containing only Fe^2+ ions)

weighing 0.2792 g was dissolved in dilute acid solution,

and all the Fe(II) was converted to Fe(III) ions.

The solution required 23.30 mL of 0.0194 M K2Cr2O7 for titration.

Calculate the percent by mass of iron in the ore.

 

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소모된 K2Cr2O7의 몰수를 계산하면,

(0.0194 mol/L) (23.30/1000 L) = 0.000452 mol K2Cr2O7

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

Cr2O7^2- + 6Fe^2+ + 14H^+ → 2Cr^3+ + 6Fe^3+ + 7H2O

 

0.000452 mol K2Cr2O7와 반응한 Fe^2+의 몰수를 계산하면,

Cr2O7^2- : Fe^2+ = 1 : 6 = 0.000452 mol : ? mol

? = 6 × 0.000452 = 0.002712 mol Fe^2+

 

 

 

0.002712 mol × (55.845 g/mol) = 0.15145 g Fe

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

(0.15145 g / 0.2792 g) × 100 = 54.2%

 

 

 

 

[키워드] Fe^2+ 02792 g Fe(lll) 00194 M K2Cr2O7 2330 mL mass