화학량론. 156 mL HSiCl3 10.0 atm 35.0℃ 93.1%

화학량론. 156 mL HSiCl3 10.0 atm 35.0℃ 93.1%

 

 

Silane, SiH4, is the silicon analogue of methane, CH4. It is prepared industrially through a two-step process. The chemical equation for the last step is:

4HSiCl3(l) → SiH4(g) + 3SiCl4(l)

When 156 mL of HSiCl3 (d = 1.34 g/mL) is heated, what volume (L) of SiH4 at 10.0 atm and 35.0℃ will be obtained if the percent yield of the reaction is 93.1%?

 

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156 mL HSiCl3의 질량을 계산하면,

156 mL × (1.34 g/mL) = 209.04 g HSiCl3

 

 

HSiCl3의 몰질량 = 135.4524 g/mol ≒ 135.45 g/mol 이므로,

209.04 g HSiCl3의 몰수를 계산하면,

209.04 g / (135.45 g/mol) = 1.5433 mol HSiCl3

( 참고 https://ywpop.tistory.com/7738 )

 

 

HSiCl3 : SiH4 = 4 : 1 계수비(= 몰수비) 이므로,

1.5433 mol HSiCl3 반응하면 생성되는 SiH4의 몰수를 계산하면,

HSiCl3 : SiH4 = 4 : 1 = 1.5433 mol : ? mol

? = 1 × 1.5433 / 4 = 0.385825 mol SiH4

---> 수율 100%인 경우

 

 

수율이 93.1%이므로,

0.385825 mol × (93.1/100) = 0.359203075 mol SiH4

 

 

PV = nRT 로부터,

( 식 설명 https://ywpop.tistory.com/3097 )

 

V = nRT / P

= [(0.359203075) (0.08206) (273.15+35)] / (10.0)

= 0.908309 L SiH4

 

 

답: 0.908 L