Calculate the amount of carbon dioxide that could be produced when

(i) 1 mole of carbon is burnt in air.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

1. 1 mole carbon = 1 x12 = 12gm

  C  +  O₂  →  CO₂

  12  32

Since the amount of carbon is less than oxygen therefore carbon is the limiting reagent therefore

  12 gram of carbon = 44 gram of CO₂

1. From the above equation, 12 gram carbon require 32 gram of oxygen, but only 16 gram of oxygen is available, therefore oxygen is the limiting reagent.

  32 gram of dioxygen = 44 gram of CO₂

Therefore 16 gm of dioxygen = 44 x 16 / 32  = 22 gm of CO₂

1. Again the amount of carbon is present in excess but dioxygen is less than the required amount as per equation therefore dioxygen is the limiting reagent.

  32 gram of dioxygen = 44 gram of CO₂

Therefore 16 gm of dioxygen = 44 x 16 / 32  = 22 gm of CO₂