0.105 M CaCl2 37.2 mL 0.125 M Na2CO3 mL

0.105 M CaCl2 37.2 mL 0.125 M Na2CO3 mL

 

 

0.105 M CaCl2 용액 37.2 mL에 들어있는

칼슘 이온을 침전시키는데 필요한

0.125 M 탄산 소듐 용액은 몇 mL인가?

 

 

How many mL of 0.125 M sodium carbonate solution

would be needed to completely react with

37.2 mL of 0.105 M CaCl2 solution?

Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2NaCl(aq)

 

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(0.105 mol/L) × (37.2/1000 L) = 0.003906 mol CaCl2

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

CaCl2(aq) + Na2CO3(aq) → CaCO3(s) + 2NaCl(aq)

 

CaCl2 : Na2CO3 = 1 : 1 계수비(= 몰수비) 이므로,

필요한 Na2CO3의 몰수 = 0.003906 mol

 

 

 

0.003906 mol / (0.125 mol/L) = 0.031248 L

= 31.248 mL Na2CO3

 

 

 

 

또는

MV = M’V’

( 참고 https://ywpop.tistory.com/4689 )

 

(0.105 M) (3.72 mL) = (0.125 M) (? mL)

 

? = (0.105) (37.2) / (0.125) = 31.248 mL

 

 

 

답: 31.2 mL

 

 

 

 

[ 관련 예제 https://ywpop.tistory.com/14006 ] 0.125 M CaCl2 용액 29.06 mL와 Na2CO3 용액 35.00 mL 반응

 

 

 

[키워드] CaCl2 + Na2CO3 + CaCl2