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 mole carbon = 1 x12 = 12gm

C + O_{2} → CO_{2}

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 CO_{2}

- 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 CO_{2}

Therefore 16 gm of dioxygen = 44 x 16 / 32 = 22 gm of CO_{2}

- 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 CO_{2}

Therefore 16 gm of dioxygen = 44 x 16 / 32 = 22 gm of CO_{2}

**
**