Nâ‚‚ : limiting reactant
Hâ‚‚ : excess reactant
Further explanation
Given
mass of Nâ‚‚ = 100 g
mass of Hâ‚‚ = 100 g
Required
Limiting reactant
Excess reactant
Solution
Reaction
N₂+3H₂⇒2NH₃
mol Nâ‚‚(MW=28 g/mol) :
[tex]\tt mol=\dfrac{mass}{MW}=\dfrac{100}{28}=3.571[/tex]
mol Hâ‚‚(MW= 2 g/mol) :
[tex]\tt mol=\dfrac{100}{2}=50[/tex]
A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients, and small or exhausted reactans become a limiting reactants
From the equation, mol ratio Nâ‚‚ : Hâ‚‚ = 1 : 3, so :
[tex]\tt \dfrac{3.571}{1}\div \dfrac{50}{3}=3.571\div 16.6[/tex]
Nâ‚‚ becomes a limiting reactant (smaller ratio) and Hâ‚‚ is the excess reactant
