heat pack 4Fe(s) + 3O2(g) → 2Fe2O3(s) ... ΔH = –1652 kJ

heat pack 4Fe(s) + 3O2(g) → 2Fe2O3(s) ... ΔH = –1652 kJ

4.00 mol Fe 1.00 mol Fe2O3

 

 

상업용 열주머니(heat pack)에서의 반응은 다음과 같다.

 

 

 

The overall reaction in a commercial heat pack can be represented as

4Fe(s) + 3O2(g) → 2Fe2O3(s) ... ΔH = –1652 kJ

a) How much heat is released when 4.00 mol iron is reacted with excess O2?

b) How much heat is released when 1.00 mol Fe2O3 is produced?

c) How much heat is released when 1.00 g iron is reacted with excess O2?

d) How much heat is released when 10.0 g Fe and 2.00 g O2 are reacted?

 

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

 

a) How much heat is released when 4.00 mol iron is reacted with excess O2?

4Fe(s) + 3O2(g) → 2Fe2O3(s) ... ΔH = –1652 kJ

---> 4Fe = 4.00 mol Fe이므로,

 

답: ΔH = –1652 kJ

 

 

 

 

b) How much heat is released when 1.00 mol Fe2O3 is produced?

4Fe(s) + 3O2(g) → 2Fe2O3(s) ... ΔH = –1652 kJ

---> 2 mol Fe2O3(s) 생성될 때 ΔH = –1652 kJ 이므로,

(–1652 kJ) / (2 mol) = –826 kJ/mol

 

답: ΔH = –826 kJ

 

 

 

 

c) How much heat is released when 1.00 g iron is reacted with excess O2?

Fe의 몰질량 = 55.845 g/mol 이므로,

1.00 g / (55.845 g/mol) = 0.0179 mol Fe

( 참고 https://ywpop.tistory.com/7738 )

 

 

(–1652 kJ) / (4 mol) × (0.0179 mol) = –7.39 kJ

 

답: ΔH = –7.39 kJ

 

 

 

 

d) How much heat is released when 10.0 g Fe and 2.00 g O2 are reacted?

> 10.0 g / (55.845 g/mol) = 0.179 mol Fe

> 2.00 g / (32.00 g/mol) = 0.0625 mol O2

 

 

4Fe(s) + 3O2(g) → 2Fe2O3(s)

 

Fe : O2 = 4 : 2 = 0.179 mol : ? mol

 

? = 2 × 0.179 / 4 = 0.0895 mol O2

---> O2는 이만큼 없다, 부족하다?

---> O2 = 한계 반응물.

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

 

 

4Fe(s) + 3O2(g) → 2Fe2O3(s) ... ΔH = –1652 kJ

 

(–1652 kJ) / (3 mol) × (0.0625 mol) = –34.4 kJ

 

답: ΔH = –34.4 kJ

 

 

 

 

[키워드] heat pack 기준문서, heat pack dic