0.600 mol M reacts with fluorine to form 46.8 g MF2

0.600 mol M reacts with fluorine to form 46.8 g MF2

 

 

금속 M 시료 0.600 mol이 과량의 플루오린과 반응을 완결하여

MF2 46.8 g을 생성했을 때 생성물의 F 원자의 몰수와 M의 질량

 

 

A sample of 0.600 mol of a metal, M, reacts completely with excess fluorine to form 46.8 g of MF2.

a. How many moles of Fluorine are in the sample of MF2 that forms?

b. How many moles of M are in the sample of MF2 that forms?

c. What is the element that is represented by M in this compound?

 

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a.

M : F = 1 : 2 = 0.600 mol : ? mol

? = 2 × 0.600 = 1.20 mol F

 

 

b.

F의 몰질량 = 19.00 g/mol

n = W / M ---> W = n × M

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

 

M의 질량 = MF2의 질량 – F2의 질량

= 46.8 g - [1.20 mol × (19.00 g/mol)]

= 24.0 g M

 

 

c.

몰질량 = g / mol

= 24.0 g / 0.600 mol

= 40.0 g/mol

---> 칼슘, Ca