42.50 mL of 12.0 M nitric acid react with silver

42.50 mL of 12.0 M nitric acid react with silver

 

 

Consider the reaction between silver and nitric acid for which the unbalanced equation is

Ag(s) + H^+(aq) + NO3^-(aq) → Ag^+(aq) + NO2(g) + H2O

a) Balance the equation.

b) If 42.50 mL of 12.0 M nitric acid furnishes enough H^+ to react with silver, how many grams of silver react?

 

 

은과 질산의 다음 반응에 대하여 생각하자.

Ag(s) + 2H^+(aq) + NO3^-(aq) → Ag^+(aq) + NO2(g) + H2O

42.50 mL의 12.0 M 질산이 은과 반응하는데 충분한 H^+를 제공해준다고 하면,

몇 그램의 은이 반응하겠는가?

 

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a) Balance the equation.

Ag(s) + 2H^+(aq) + NO3^-(aq) → Ag^+(aq) + NO2(g) + H2O

( 설명 https://ywpop.tistory.com/13805 )

 

 

 

b) If 42.50 mL of 12.0 M nitric acid furnishes enough H^+ to react with silver, how many grams of silver react?

 

(12.0 mol/L) × (42.50/1000) L = 0.510 mol HNO3

 

 

HNO3(aq) → H^+(aq) + NO3^-(aq)

HNO3 : H^+ = 1 : 1 계수비(= 몰수비) 이므로,

HNO3의 몰수 = H^+의 몰수 = 0.510 mol

 

 

Ag(s) + 2H^+(aq) + NO3^-(aq) → Ag^+(aq) + NO2(g) + H2O

Ag : H^+ = 1 : 2 계수비(= 몰수비) 이므로,

0.510 mol H^+와 반응하는 Ag의 몰수를 계산하면,

Ag : H^+ = 1 : 2 = ? mol : 0.510 mol

? = 0.510 / 2 = 0.255 mol Ag

 

 

Ag의 몰질량 = 107.87 g/mol

0.255 mol × (107.87 g/mol) = 27.5 g Ag

( 계산 설명 https://ywpop.tistory.com/7738 )

 

 

답: 27.5 g