150.0 g WCl6 228 g WCl6 175 g Bi W6Cl12

150.0 g WCl6 228 g WCl6 175 g Bi W6Cl12

 

 

The reaction of tungsten hexachloride (WCl6) with bismuth

gives hexatungsten dodecachloride (W6Cl12).

WCl6 + Bi → W6Cl12 + BiCl3 Unbalanced

a) Balance the equation.

b) How many grams of bismuth react with 150.0 g of WCl6?

c) When 228 g of WCl6 react with 175 g of Bi,

how much W6Cl12 is formed based on the limiting reactant?

 

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a) Balance the equation.

 

6WCl6 + 8Bi → W6Cl12 + 8BiCl3

 

 

 

 

b) How many grams of bismuth react with 150.0 g of WCl6?

 

WCl6의 몰질량 = 396.56 g/mol 이므로,

150.0 g / (396.56 g/mol) = 0.37825 mol WCl6

( 참고 https://ywpop.tistory.com/7738 )

 

 

0.37825 mol WCl6와 반응하는 Bi의 몰수를 계산하면,

WCl6 : Bi = 6 : 8 = 0.37825 mol : ? mol

 

? = 8 × 0.37825 / 6 = 0.50433 mol Bi

 

 

Bi의 몰질량 = 208.98 g/mol 이므로,

0.50433 mol × (208.98 g/mol) = 105.4 g Bi

 

 

 

 

c) When 228 g of WCl6 react with 175 g of Bi,

how much W6Cl12 is formed based on the limiting reactant?

 

> 228 g / (396.56 g/mol) = 0.5749 mol WCl6

> 175 / 208.98 = 0.8374 mol Bi

 

 

0.5749 mol WCl6와 반응하는 Bi의 몰수를 계산하면,

WCl6 : Bi = 6 : 8 = 0.5749 mol : ? mol

 

? = 8 × 0.5749 / 6 = 0.7665 mol Bi

---> Bi는 이만큼 있다, 남는다.

---> Bi는 과잉 반응물.

---> WCl6가 한계 반응물.

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

 

 

0.5749 mol WCl6 반응 시 생성되는 W6Cl12의 몰수를 계산하면,

WCl6 : W6Cl12 = 6 : 1 = 0.5749 mol : ? mol

 

? = 0.5749 / 6 = 0.09582 mol W6Cl12

 

 

W6Cl12의 몰질량 = 1528.48 g/mol 이므로,

0.09582 × 1528.48 = 146.5 g W6Cl12

 

 

 

 

[키워드] 육염화텅스텐 기준문서, 십이염화헥사텅스텐 기준문서, W6Cl12 dic, 228 g WCl6 175 g Bi gram W6Cl12

 

[FAQ] [①23/10/04]