Question

If 100.0g of nitrogen is reacted with 100.0g of hydrogen, what is the excess reactant? What is the limiting reactant? Show your work.

If someone could just help me understand how to do this, it would be much appreciated. (:

Answer 1

N₂ : limiting reactant

H₂ : excess reactant

Further explanation

Given

mass of N₂ = 100 g

mass of H₂ = 100 g

Required

Limiting reactant

Excess reactant

Solution

Reaction

N₂+3H₂⇒2NH₃

mol N₂(MW=28 g/mol) :

`\tt mol=(mass)/(MW)=(100)/(28)=3.571`

mol H₂(MW= 2 g/mol) :

`\tt mol=(100)/(2)=50`

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 N₂ : H₂ = 1 : 3, so :

`\tt (3.571)/(1)/ (50)/(3)=3.571/ 16.6`

N₂ becomes a limiting reactant (smaller ratio) and H₂ is the excess reactant