0.105 M Fe(NO3)3 75.0 mL 0.150 M NaOH 125 mL Fe(OH)3 mass

0.105 M Fe(NO3)3 75.0 mL 0.150 M NaOH 125 mL Fe(OH)3 mass

 

 

What mass of Fe(OH)3 would be produced by reacting

75.0 mL of 0.105 M Fe(NO3)3 with 125 mL of 0.150 M NaOH?

 

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각 반응물의 몰수를 계산하면,

(0.105 mol/L) (75.0/1000 L) = 0.007875 mol Fe(NO3)3

( 참고 https://ywpop.tistory.com/7787 )

 

(0.150 mol/L) (125/1000 L) = 0.01875 mol NaOH

 

 

 

균형 맞춘 화학 반응식

Fe(NO3)3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaNO3(aq)

( 참고 https://ywpop.tistory.com/11299 )

 

Fe(NO3)3 : NaOH = 1 : 3 계수비(= 몰수비) 이므로,

0.007875 mol Fe(NO3)3와 반응하는 NaOH의 몰수를 계산하면,

Fe(NO3)3 : NaOH = 1 : 3 = 0.007875 mol : ? mol

 

? = 3 × 0.007875 = 0.023625 mol NaOH

---> NaOH는 이만큼 없다, 부족하다?

---> NaOH = 한계 반응물.

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

 

 

 

NaOH : Fe(OH)3 = 3 : 1 = 0.01875 mol : ? mol

 

? = 0.01875 / 3 = 0.00625 mol Fe(OH)3

 

 

 

Fe(OH)3의 몰질량 = 106.87 g/mol 이므로,

0.00625 mol × (106.87 g/mol) = 0.668 g Fe(OH)3

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

답: 0.668 g

 

 

 

 

[FAQ] [①23/04/11]