4Au + 8KCN + O2 + 2H2O Au 29.0 g 1 ounce

4Au + 8KCN + O2 + 2H2O Au 29.0 g 1 ounce

 

 

For many years the extraction of gold

- that is, the separation of gold from other materials -

involved the use of potassium cyanide:

4Au + 8KCN + O2 + 2H2O → 4KAu(CN)2 + 4KOH

What is the minimum amount of KCN in moles

needed to extract 29.0 g (about an ounce) of gold?

 

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Au의 몰질량 = 196.97 g/mol 이므로,

29.0 g / (196.97 g/mol) = 0.147 mol Au

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

Au : KCN = 1 : 2 계수비(= 몰수비) 이므로,

0.147 mol Au와 화학량론적으로 반응하는 KCN의 몰수를 계산하면,

Au : KCN = 1 : 2 = 0.147 mol : ? mol

? = 2 × 0.147 = 0.294 mol KCN

 

 

 

답: 0.294 mol