500.0 g Al + 500.0 g Fe2O3 produces 166.5 g Fe

500.0 g Al + 500.0 g Fe2O3 produces 166.5 g Fe

 

 

The reaction of finely divided aluminum and iron(III) oxide, Fe2O3, is called the thermite reaction. It produces a tremendous amount of heat, making the welding of railroad track possible. The reaction of 500.0 grams of aluminum and 500.0 grams of iron(III) oxide produces 166.5 grams of iron.

Fe2O3 + 2Al → 2Fe + Al2O3 + Heat

a. Calculate the mass of iron that should be released by this reaction.

b. What is the percent yield of iron?

 

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Al의 몰질량 = 26.98 g/mol 이므로,

500.0 g Al의 몰수를 계산하면,

500.0 g / (26.98 g/mol) = 18.53 mol Al

( 참고 https://ywpop.tistory.com/7738 )

 

 

Fe2O3의 몰질량 = 159.69 g/mol 이므로,

500.0 g / (159.69 g/mol) = 3.131 mol Fe2O3

 

 

Fe2O3 : Al = 1 : 2 계수비(= 몰수비) 이므로,

Fe2O3 : Al = 1 : 2 = 3.131 mol : ? mol

? = 2 × 3.131 = 6.262 mol Al

---> Al은 이만큼 있다. 반응하고도 남는다.

---> Al은 과잉 반응물. 따라서 Fe2O3가 한계 반응물.

 

 

Fe2O3 : Fe = 1 : 2 = 3.131 mol : ? mol

? = 2 × 3.131 = 6.262 mol Fe

 

 

Fe의 몰질량 = 55.85 g/mol 이므로,

6.262 mol Fe의 몰수를 계산하면,

6.262 mol × (55.85 g/mol) = 349.7 g Fe

 

 

(166.5 g / 349.7 g) × 100 = 47.61%