Question

How many moles of butane (C4H10(g)) were combusted in the presence of excess

oxygen gas if 55.8 L of water vapour at STP is collected?

Answer 1

Moles of Butane combusted : 0.4982

Further explanation

Given

55.8 water vapour at STP

Required

moles of Butane

Solution

Reaction(Combustion of Butane) :

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

1 mol at STP = 22.4 L, so mol of H₂O for 55.8 L :

=55.8 : 22.4

=2.491 moles

mol H₂O based on Butane as a limiting reactant(excess Oxygen)

From the equation, mol ratio Butane-C₄H₁₀ : H₂O = 2 : 10, so mol Butane :

`\tt =(2)/(10)* 2.491\\\\=0.4982`