0.250 M AlCl3 10.0 mL gram solid NaOH

0.250 M AlCl3 10.0 mL gram solid NaOH

 

 

0.250 M AlCl3 10.0 mL에 있는 모든 알루미늄을 침전시키기 위해

몇 g의 고체 NaOH를 첨가해야 하는가?

 

 

How many grams of solid NaOH must be added to

10.0 mL of 0.250 M AlCl3 to exactly precipitate all of the aluminum?

 

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(0.250 mol/L) (10.0/1000 L) = 0.00250 mol AlCl3

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

AlCl3 + 3NaOH → Al(OH)3 + 3NaCl

 

AlCl3 : NaOH = 1 : 3 계수비(= 몰수비) 이므로,

0.00250 mol AlCl3와 반응하는 NaOH의 몰수를 계산하면,

AlCl3 : NaOH = 1 : 3 = 0.00250 mol : ? mol

? = 3 × 0.00250 = 0.00750 mol NaOH

 

 

 

NaOH의 몰질량 = 40.00 g/mol

(0.00750 mol) (40.00 g/mol) = 0.300 g NaOH

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

답: 0.300 g