CaF2 6.00 kg excess H2SO4 HF 2.86 kg yield percent

CaF2 6.00 kg excess H2SO4 HF 2.86 kg yield percent

 

 

In the following reaction,

6.00 kg of CaF2 is treated with an excess of H2SO4,

and produces 2.86 kg of HF.

Calculate the percent yield of HF.

CaF2 + H2SO4 → CaSO4 + 2HF

 

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> CaF2의 몰질량 = 78.07 g/mol

> HF의 몰질량 = 20.01 g/mol

 

 

 

6.00 kg / (78.07 kg/kmol) = 0.076854 kmol CaF2

 

 

 

CaF2 + H2SO4 → CaSO4 + 2HF

CaF2 : HF = 1 : 2 = 0.076854 kmol : ? kmol

 

? = 2 × 0.076854 = 0.153708 kmol HF

 

 

 

0.153708 kmol × (20.01 kg/kmol) = 3.075697 kg HF

---> 이론적 수득량

 

 

 

HF의 수득률을 계산하면,

(2.86 kg / 3.075697 kg) × 100 = 92.99%

( 참고 https://ywpop.tistory.com/61 )

 

 

 

답: 93.0%