255 g Al 535 g Cl2 gas 300 g AlCl3 percent yield

255 g Al 535 g Cl2 gas 300 g AlCl3 percent yield

 

 

Aluminum metal reacts with chlorine gas

to produce aluminum chloride.

In one preparation, 255 g of aluminum is placed

in a container holding 535 g of chlorine gas.

After reaction ceases, it is found that

300. g of aluminum chloride has been produced.

a) Write the balanced equation for the reaction.

b) What mass of aluminum chloride

can be produced by these reactants?

c) What is the percentage yield of aluminum chloride?

 

---------------------------------------------------

 

a) Write the balanced equation for the reaction.

 

2Al(s) + 3Cl2(g) → 2AlCl3(s)

 

2Al + 3Cl2 → 2AlCl3

 

 

 

 

b) What mass of aluminum chloride

can be produced by these reactants?

 

Al의 몰질량 = 26.98 g/mol 이므로,

255 g / (26.98 g/mol) = 9.451 mol Al

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

Cl2의 몰질량 = 70.91 g/mol 이므로,

535 / 70.91 = 7.545 mol Cl2

 

 

 

2Al + 3Cl2 → 2AlCl3

 

Al : Cl2 = 2 : 3 계수비(= 몰수비) 이므로,

9.451 mol Al과 반응하는 Cl2의 몰수를 계산하면,

Al : Cl2 = 2 : 3 = 9.451 mol : ? mol

 

? = 3 × 9.451 / 2 = 14.18 mol Cl2

---> Cl2는 이만큼 없다, 부족하다?

---> Cl2 = 한계 반응물, Al = 과잉 반응물.

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

 

 

 

Cl2 : AlCl3 = 3 : 2 계수비(= 개수비) 이므로,

Cl2 : AlCl3 = 3 : 2 = 7.545 mol = ? mol

 

? = 2 × 7.545 / 3 = 5.03 mol AlCl3

( 생성물의 양은 한계 반응물에 의해 결정된다. )

 

 

 

AlCl3의 몰질량 = 133.34 g/mol 이므로,

5.03 mol × (133.34 g/mol) = 671 g AlCl3

---> 이론적 수득량

 

 

 

 

c) What is the percentage yield of aluminum chloride?

 

수득률 = (실제 수득량 / 이론적 수득량) × 100

( 참고 https://ywpop.tistory.com/61 )

 

= (300. g / 671 g) × 100

= 44.7%

 

 

 

 

[키워드] 한계 반응물 기준문서, 수득률 기준문서