Question

identify the limiting reagent in a reaction of 12 gm C with 64 gm O2?​

Answer 1

Answer:

Explanation:

24g

2mol

2C(s)

​

 

​

+  

32g

1mol

O  

2

​

 

​

 

​

→  

56g

2mol

2CO(g)

​

 

​

 

Let carbon be completely consumed.

24g carbon give 56 g CO.

Let O  

2

​

 is completely consumed.  

∵ 32 g O  

2

​

 give 56 g CO.

∴ 96 g O  

2

​

 Will give  

32

56

​

×96gCO=168gCO

Since, carbon gives least amount of product, te.,56 g CO or 2 mole CO, hence carbon will be the limiting reactant.  

∴ Excess reactant is O  

2

​

.  

Amount of O  

2

​

 used =56−24=32g  

Amount of O  

2

​

 left =96−32=64g

32g O  

2

​

 react with 24 g carbon  

∴ 96 g O  

2

​

 will react with 72g carbon.  

Thus, carbon should be taken 72g so that nothing is left at the end of the reaction.