Question

Consider the chemical equation.

4NH3+5O2⟶4NO+6H2O4NH3+5O2⟶4NO+6H2O

When 26 g of ammonia burn in 26 g of oxygen, what is the limiting reagent?

NH3NH3 and O2O2

NONO

O2O2

H2OH2O

NH3NH3

Answer 1

Balanced reaction is : 4 NH₃ + 5 O₂ 4 NO + 6 H₂O

Case 1 : Let NH₃ be the limiting reagent

mass NH₃ = 26 g

moles NH₃ = (mass NH₃) / (molar mass NH₃)

moles NH₃ = (26 g) / (17.0 g/mol)

moles NH₃ = 1.53 mol

moles NO formed = moles NH₃

moles NO formed = 1.53 mol

Case 2 : let O₂ be the limiting reagent

mass O₂ = 26 g

moles O₂ = (mass O₂) / (molar mass O₂)

moles O₂ = (26 g) / (32.0 g/mol)

moles O₂ = 0.8125 mol

moles NO formed = (moles O₂ consumed) * (4 moles NO / 5 moles O₂)

moles NO formed = (0.8125 mol) * (4/5)

moles NO formed = (0.8125 mol) * (0.8)

moles NO formed = 0.650 mol

Since less moles of NO are formed in Case 2, therefore, O₂ is the limiting reagent.