PCl3 4.00×10^23 PCl5 0.250 mol Cl gram P mol

PCl3 4.00×10^23 PCl5 0.250 mol Cl gram P mol

 

 

A mixture contains 4.00×10^23 molecules of PCl3,

and 0.250 mol of PCl5.

a. How many grams of Cl are present in the mixture?

b. How many moles of P are in the mixture?

 

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> Cl의 몰질량 = 35.45 g/mol

 

 

 

PCl3의 몰질량 = 137.33 g/mol 이므로,

(4.00×10^23 개) (1 mol / 6.022×10^23 개) (137.33 g/mol)

= (4.00×10^23) (1 / (6.022×10^23)) (137.33)

= 91.22 g PCl3

( 참고: 개수를 질량으로 https://ywpop.tistory.com/23528 )

 

 

 

PCl3 : 3Cl = 137.33 : 3(35.45) = 91.22 g : ? g

 

? = 3(35.45) × 91.22 / 137.33 = 70.64 g Cl

( 참고 https://ywpop.tistory.com/23533 )

 

 

 

PCl5의 몰질량 = 208.24 g/mol 이므로,

0.250 mol × (208.24 g/mol) = 52.06 g PCl5

 

 

 

PCl5 : 5Cl = 208.24 : 5(35.45) = 52.06 g : ? g

 

? = 5(35.45) × 52.06 / 208.24 = 44.31 g Cl

 

 

 

70.64 + 44.31 = 115 g Cl

---> a) 답

 

 

 

 

(4.00×10^23 개) (1 mol / 6.022×10^23 개)

= (4.00×10^23) (1 / (6.022×10^23))

= 0.664 mol PCl3

 

 

 

PCl3, PCl5, 둘 다,

분자의 몰수 = P 원자의 몰수 이므로,

0.664 + 0.250 = 0.914 mol P

---> b) 답