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?

Volume CO₂ produced  := 33.85 L

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

answer: the primary difference between a hypothesis and a theory is that b hypothesis is falsifiable, but a theory has been proven to be correct. theory is an explanation of something that is happening but hypothesis is a guess of something that can happen

explanation:

answer:

the reaction is type of combination reaction.

explanation:

on the product side, there are 2 molecules of bonded to fe   which means one atom of fe are combined two molecules of . so, balanced chemical equation becomes :

this reaction follows the reaction type of combination reaction. a combination reaction is defined as chemical reaction in which reactants combines together to form a single product.