If 6.0g of carbon is heated in air the mass of the product obtained could be either 22.0g or 14.0g depending on the amount of air present. With what law would this be in accordance (C=12,O

In accordance with Dalton's Law of multiple proportions

Given

6.0g of carbon

22.0g or 14.0g of product

Required

related laws

Solution

the amount of air present ⇒ as an excess or limiting reactant

C+0.5O₂⇒CO

6 + 8 = 14 g

mol O₂=8 g : 32 g/mol=0.25

mol C = 6 g : 12 g/mol = 0.5(2 x mol O₂)

mol CO= 2 x mol O₂ = 0.5 mol = 0.5 x 28 g/mol = 14 g

C+O₂⇒CO₂

6 + 16 = 22 g

mol C = 6 g : 12 g/mol = 0.5

mol O₂ = 16 g : 32 g/mol=0.5

mol CO₂ = 22 g : 44 g/mol = 0.5

if the mass firs element (C) constant, then the mass of the second element(O) in the two compounds will have a ratio as a simple integer.

CO = 6 : 8

CO₂ = 6 : 16

the ratio O = 8 : 16 = 1 : 2

In accordance with Dalton's Law of multiple proportions