# 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

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TELL WITH PROPER SOLUTION PLS

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C+O2 --> CO2(12g of C requires 32g of O2 to give 44g of CO2)

i)Air contains excess of O2 and 1 mole of C is present so it is limiting and thus gives 1 mole of CO2

ii)we have 1 mole of C but only 16g of Oagainst the required 32g(hence limiting) therefore we get only 1/2 mole of CO2

iii)again O2 continues to be limiting and thus again we get only 1/2 mole of CO2

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