AlBr3 Al 3.13 g Br2 28.8 g

AlBr3 Al 3.13 g Br2 28.8 g

 

 

AlBr3를 만들기 위하여 Al 3.13 g과 Br2 28.8 g을 섞었다.

 

 

A chemist mixed 3.13 g of aluminium with 28.8 g of bromine

to form aluminium bromide according to the equation:

2Al + 3Br2 → 2AlBr3

 

a) Identify the limiting reagent.

b) What is the maximum mass of aluminium bromide which can be made?

c) With respect to the reactant that is in excess,

what mass remains after the reaction is completed?

 

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3.13 g / (26.98 g/mol) = 0.116 mol Al

( 참고: n=W/M https://ywpop.tistory.com/7738 )

 

 

28.8 g / (159.81 g/mol) = 0.180 mol Br2

 

 

 

Al : Br2 = 2 : 3 계수비(= 몰수비) 이므로,

0.116 mol Al과 반응하는 Br2의 몰수를 계산하면,

Al : Br2 = 2 : 3 = 0.116 mol : ? mol

? = 3 × 0.116 / 2 = 0.174 mol

---> Br2는 이만큼 있다, 남는다?

---> Br2는 과잉 반응물.

---> 따라서 Al이 한계 반응물.

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

 

 

 

 

Al : AlBr3 = 2 : 2 계수비(= 몰수비) 이므로,

생성물 AlBr3의 몰수 = 0.116 mol

 

 

 

AlBr3의 몰질량 = 266.69 g/mol 이므로,

0.116 mol × (266.69 g/mol) = 30.9 g AlBr3

 

 

 

 

0.180 mol - 0.174 mol = 0.006 mol Br2

0.006 mol × (159.81 g/mol) = 0.959 g Br2