MnI2 + F2 반응. 1.36 g MnI2 + 37.1 g fluorine

MnI2 + F2 반응. 1.36 g MnI2 + 37.1 g fluorine

 

 

Manganese(II) iodide react with fluorine to produce manganese(III) fluoride and iodine pentafluoride.

 

 

Manganese(III) fluoride can be formed by the reaction of manganese(II) iodide with fluorine.

 

 

1.36 g MnI2 react with 37.1 g fluorine

한계 반응물은? 질량은?(어떤 질량?)

 

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균형 맞춘 화학 반응식

2MnI2 + 13F2 → 2MnF3 + 4IF5

 

 

 

MnI2의 몰질량 = 308.75 g/mol

1.36 g / (308.75 g/mol) = 0.00440486 mol MnI2

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

F2의 몰질량 = 38.00 g/mol

37.1 / 38.00 = 0.976316 mol F2

 

 

 

2MnI2 + 13F2 → 2MnF3 + 4IF5

0.00440486 mol MnI2와 반응하는 F2의 몰수를 계산하면,

MnI2 : F2 = 2 : 13 = 0.00440486 mol : ? mol

? = 13 × 0.00440486 / 2 = 0.0286316 mol F2

---> F2는 이만큼 있다, 남는다.

---> F2는 과잉, MnI2가 한계 반응물.

( 참고: 한계 반응물 https://ywpop.tistory.com/3318 )