OBr^- 1.00 mL NH3 1.69 mg

OBr^- 1.00 mL NH3 1.69 mg

 

 

암모니아는 OBr^-와 다음과 같이 반응한다.

2NH3 + 3OBr^- → N2 + 3Br^- + 3H2O

OBr^- 용액 1.00 mL가 NH3 1.69 mg과 반응하면

OBr^- 용액의 몰농도는 얼마인가?

 

 

Ammonia reacts with hypobromite, OBr^- as follows:

2NH3 + 3OBr^- → N2 + 3Br^- + 3H2O.

What is the molarity of a hypobromite solution

if 1.00 mL of the OBr^- solution

reacts with 1.69 mg of NH3 (MW 17.03)?

 

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1.69 mg / (17.03 mg/mmol) = 0.0992 mmol NH3

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

2NH3 + 3OBr^- → N2 + 3Br^- + 3H2O

 

NH3 : OBr^- = 2 : 3 = 0.0992 mmol : ? mmol

 

? = 3 × 0.0992 / 2 = 0.149 mmol OBr^-

 

 

 

0.149 mmol / 1.00 mL = 0.149 mmol/mL

( 참고 https://ywpop.tistory.com/3222 )

 

 

 

답: 0.149 M