Question

Cigarette lighters use butane, or C4H10, as their fuel. If 120.0 grams of butane burn with only 55.0 liters of oxygen gas at STP, how many liters of carbon dioxide are produced? Which reactant is your limiting reactant?

Answer 1

Volume CO₂ produced := 33.85 L

Further explanation

Given

120.0 grams of butane

55.0 liters of oxygen

Required

Carbon dioxide produced

Solution

Reaction(combustion of Butane) :

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O.

mol of Butane (MW=58 g/mol) :

mol = mass : MW

mol = 120 : 58

mol = 2.069

At STP(1 atm, 273 K) , 1 mol gas=22.4 L, so for 55 L Oxygen :

mol O₂= 55 : 22.4 =2.455

Mol ratio of Butane and Oxygen = mol : reaction coefficient :

C₄H₁₀ : O₂ = 2.069/2 : 2.455/13 = 1.0345 : 0.189

Limiting reactant ⇒ O₂ (smaller ratio)

So mol CO₂ based on limiting reactant(O₂)

From equation, mol CO₂ : O₂ = 8 : 13, so mol CO₂ :

= 8/13 x mol O₂

= 8/13 x 2.455

= 1.511

Volume CO₂ at STP :

= 1.511 x 22.4 L

= 33.85 L