In accordance with Dalton's Law of multiple proportions
<h3>Further explanation</h3>
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
- air(O₂) as a limiting reactant(product=14 g)
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
- air(O₂) as an excess reactant(product=22 g) an C as a limiting reactant
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
