4Fe(s) + 3O2(g) → 2Fe2O3(s) ΔH = –1.7×10^3 kJ Fe 2.00 g 150 kJ

4Fe(s) + 3O2(g) → 2Fe2O3(s) ΔH = –1.7×10^3 kJ Fe 2.00 g 150 kJ

 

 

4Fe(s) + 3O2(g) → 2Fe2O3(s) ... ΔH = –1.7×10^3 kJ

a) Fe 2.00 g 반응할 때 몇 kJ 방출?

b) 150 kJ 방출될 때 녹 몇 g 형성?

 

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a) Fe 2.00 g 반응할 때 몇 kJ 방출?

Fe의 몰질량 = 55.85 g/mol 이므로,

2.00 g / (55.85 g/mol) = 0.0358 mol Fe

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

0.0358 mol × ((–1.7×10^3 kJ) / 4 mol)

= 0.0358 × ((–1.7×10^3) / 4)

= –15.2 kJ

---> 답: 15.2 kJ

 

 

 

 

b) 150 kJ 방출될 때 녹 몇 g 형성?

Fe2O3의 몰질량 = 159.69 g/mol 이므로,

 

–150 kJ × (2 mol / (–1.7×10^3 kJ)) × (159.69 g / 1 mol)

= –150 × (2 / (–1.7×10^3)) × (159.69)

= 28.2 g Fe2O3

---> 답: 28.2 g