Question

11.2 L of O2 at STP and 8 g of calcium are allowed to react. The limiting reagent in the reaction and the
unreacted volume/weight of the reactant in excess is
1) Calcium, 0.896 L of O2 2) 02,2 g of 3) Calcium, 10 L of O2 4) Calcium, 8.96 L of O2
at STP
calcium
at STP
at STP​

Answer 1

Answer:

2C(s)+  32g  1mol  O  2 →  56g  2mol  2CO(g)  

Let carbon be completely consumed.

24g carbon give 56 g CO.

Let O2  is completely consumed.  

∵ 32 g O2 give 56 g CO.

∴ 96 g O2  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 O2.

Amount of O2 used =56−24=32g  

Amount of O2 left =96−32=64g

32g O2  react with 24 g carbon  

∴ 96 g O2 will react with 72g carbon.  

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

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