화학량론. mercury in a 1.0451-g sample

화학량론. mercury in a 1.0451-g sample

 

 

The mercury in a 1.0451-g sample was precipitated with an excess of paraperiodic acid, H5IO6:

5Hg^2+ + 2H5IO6 → Hg5(IO6)2 + 10H^+

The precipitate was filtered, washed free of precipitating agent, dried, and weighed, and 0.5718 g was recovered. Calculate the percentage of Hg2Cl2 in the sample.

 

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침전된 0.5718 g Hg5(IO6)2의 몰수를 계산하면,

Hg5(IO6)2의 몰질량 = 1448.75 g/mol

0.5718 g / (1448.75 g/mol) = 0.00039469 mol Hg5(IO6)2

( 식 설명 http://ywpop.tistory.com/7738 )

 

0.00039469 mol Hg5(IO6)2에 들어있는 Hg^2+의 몰수

= 0.00039469 mol * 5 = 0.0019735 mol Hg^2+

 

Hg2Cl2의 몰수

= 0.0019735 mol / 2 = 0.00098675 mol Hg2Cl2

 

0.00098675 mol Hg2Cl2의 질량을 계산하면,

Hg2Cl2의 몰질량 = 472.09 g/mol

0.00098675 mol * (472.09 g/mol) = 0.46583 g Hg2Cl2

 

% = (0.46583 g / 1.0451 g) * 100 = 44.573%

 

 

한 번에 계산하면,

[(침전물 질량 / 1448.75 * 5 / 2 * 472.09) / 시료 질량] * 100

= [(0.5718 / 1448.75 * 5 / 2 * 472.09) / 1.0451] * 100 = 44.572%

 

 

답: 44.57%