LiOH(s) 155 g CO2(g) mass 24 hr CO2 102 g

LiOH(s) 155 g CO2(g) mass 24 hr CO2 102 g

 

 

Suppose a lithium hydroxide canister contains 155 g of LiOH(s).

What mass of CO2(g) will the canister be able to absorb?

If it is found that after 24 hours of use

the canister has absorbed 102 g of carbon dioxide,

what percentage of its capacity has been reached?

LiOH(s) + CO2(g) → Li2CO3(s) + H2O(g)

 

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LiOH의 몰질량 = 23.95 g/mol

155 g / (23.95 g/mol) = 6.4718 mol LiOH

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

균형 맞춘 화학 반응식

2LiOH + CO2 → Li2CO3 + H2O

 

LiOH : CO2 = 2 : 1 계수비(= 몰수비) 이므로,

흡수할 수 있는 CO2의 몰수를 계산하면,

LiOH : CO2 = 2 : 1 = 6.4718 mol : ? mol

? = 6.4718 / 2 = 3.2359 mol CO2

 

 

 

CO2의 몰질량 = 44.01 g/mol

3.2359 mol × (44.01 g/mol) = 142.4 g CO2

---> 흡수할 수 있는 CO2의 질량(g)

 

 

 

(102 g / 142.4 g) × 100 = 71.6%

---> 통 용량의 71.6%까지 도달.

 

 

 

 

[키워드] LiOH(s) 155 g 기준문서