0.426 g Zn 0.235 M HCl mL

0.426 g Zn 0.235 M HCl mL

 

 

Using the reaction in Sample Problem 12.14,

how many milliliters of a 0.235 M HCl solution

will react with 0.426 g of zinc?

 

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Zn의 몰질량 = 65.38 g/mol 이므로,

0.426 g / (65.38 g/mol) = 0.006516 mol Zn

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

Zn + 2HCl → ZnCl2 + H2

 

Zn : HCl = 1 : 2 = 계수비(= 몰수비) 이므로,

0.006516 mol Zn과 반응하는 HCl의 몰수를 계산하면,

Zn : HCl = 1 : 2 = 0.006516 mol : ? mol

 

? = 2 × 0.006516 = 0.01303 mol HCl

 

 

 

0.01303 mol HCl을 포함하는

0.235 M HCl 용액의 부피를 계산하면,

0.01303 mol / (0.235 mol/L) = 0.05545 L

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

답: 55.45 mL