5.00 g Ti and 6.00 g Cl2 Ti(s) + Cl2(g) ⟶ TiCl4(s)

5.00 g Ti and 6.00 g Cl2 Ti(s) + Cl2(g) ⟶ TiCl4(s)

 

 

For the unbalanced reaction shown,

calculate the amount of excess reactant (in moles)

if you have the following amounts of reactants:

5.00 g Ti and 6.00 g Cl2.

Ti(s) + Cl2(g) ⟶ TiCl4(s)

 

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> 5.00 g / (47.87 g/mol) = 0.10445 mol Ti

( 참고 https://ywpop.tistory.com/7738 )

 

> 6.00 / 70.91 = 0.0846 mol Cl2

 

 

 

균형 맞춘 반응식

Ti(s) + 2Cl2(g) ⟶ TiCl4(s)

 

Ti : Cl2 = 1 : 2 계수비(= 몰수비) 이므로,

Ti : Cl2 = 1 : 2 = 0.10445 mol : ? mol

? = 2 × 0.10445 = 0.2089 mol Cl2

---> Cl2는 이만큼 없다? 부족하다!

---> Cl2 = 한계 반응물, Ti = 과잉 반응물.

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

 

 

 

답: 0.104 mol

 

 

 

 

[참고] 반응하고 남은 양 계산

Ti : Cl2 = 1 : 2 = ? mol : 0.0846 mol

? = 0.0846 / 2 = 0.0423 mol Ti

---> 이만큼 반응(소모).

 

0.10445 – 0.0423 = 0.06215 mol Ti