1.45 M NaI 225 mL + 0.85 M Pb(NO3)2 175 mL

1.45 M NaI 225 mL + 0.85 M Pb(NO3)2 175 mL

 

 

225 mL of a 1.45 mol/L solution of sodium iodide

is added to 175 mL of a 0.85 mol/L solution of lead(II) nitrate.

a) Write the balanced chemical equation for this reaction.

b) How many grams of solid precipitate will form in the process?

c) What is the percent yield if 57.8 g of precipitate was collected?

 

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a) Write the balanced chemical equation for this reaction.

 

NaI + Pb(NO3)2 → PbI2 + NaNO3

 

2NaI + Pb(NO3)2 → PbI2 + 2NaNO3

2NaI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2NaNO3(aq)

( 참고 https://ywpop.tistory.com/2523 )

 

 

 

 

b) How many grams of solid precipitate will form in the process?

 

(1.45 mol/L) (0.225 L) = 0.326 mol NaI

 

(0.85) (0.175) = 0.149 mol Pb(NO3)2

 

 

 

NaI : Pb(NO3)2 = 2 : 1 계수비(= 몰수비) 이므로,

0.326 mol NaI와 반응하는 Pb(NO3)2의 몰수를 계산하면,

NaI : Pb(NO3)2 = 2 : 1 = 0.326 mol : ? mol

? = 0.326 / 2 = 0.163 mol Pb(NO3)2

---> Pb(NO3)2는 이만큼 없다, 부족하다?

---> Pb(NO3)2가 한계 반응물.

( 참고: 한계 반응물 https://ywpop.tistory.com/3318 )

 

 

 

Pb(NO3)2 : PbI2(s) = 1 : 1 계수비(= 몰수비) 이므로,

0.149 mol Pb(NO3)2 반응 시

침전되는 PbI2의 몰수 = 0.149 mol

 

 

 

PbI2의 몰질량 = 461.01 g/mol 이므로,

0.149 mol × (461.01 g/mol) = 68.7 g PbI2

---> 이론적 수득량

 

 

 

 

c) What is the percent yield if 57.8 g of precipitate was collected?

 

수득률 = (57.8 g / 68.7 g) × 100 = 84.1%

( 참고 https://ywpop.tistory.com/61 )