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일반화학/[03장] 화학량론

0.2500 M AgNO3 45.0 mL NaCl mass

by 영원파란 2023. 4. 13.

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0.2500 M AgNO3 45.0 mL NaCl mass

 

 

You want to analyze a silver nitrate solution.

What mass of NaCl is needed to precipitate Ag^+ ions

from 45.0 mL of 0.2500 M AgNO3 solution?

 

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

 

(0.2500 mol/L) (45.0/1000 L) = 0.01125 mol AgNO3

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

 

AgNO3 : NaCl = 1 : 1 계수비(= 몰수비) 이므로,

AgNO3의 몰수만큼 NaCl 필요.

 

 

 

NaCl의 몰질량 = 58.44 g/mol 이므로,

0.01125 mol × (58.44 g/mol) = 0.65745 g NaCl

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

답: 0.657 g

 

 

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