Butane gas is used as the fuel in disposable lighters. It burn in oxygen to form carbon dioxide gas and water vapor. What mass of carbon dioxide is produced when 53.3 g of butane is burned in 103.9 g of oxygen

Taking into account the reaction stoichiometry and lmiting reaction, 87.91 grams of CO₂ are produced when 53.3 g of butane is burned in 103.9 g of oxygen.

In first place, the balanced reaction is:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Reaction stoichiometry

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

C₄H₁₀: 2 moles
O₂: 13 moles
CO₂: 8 moles
H₂O: 10 moles

The molar mass of the compounds is:

C₄H₁₀: 58 g/mole
O₂: 32 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

C₄H₁₀: 2 moles× 58 g/mole= 116 grams
O₂: 13 moles× 32 g/mole= 416 grams
CO₂: 8 moles× 44 g/mole= 352 grams
H₂O: 10 moles× 18 g/mole= 180 grams

Limiting reagent

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction a simple rule of three as follows: if by stoichiometry 116 grams of C₄H₁₀ reacts with 416 grams of O₂, if 53.3 grams of C₄H₁₀ react how much mass of O₂ will be needed?

[tex]mass of O_{2} =\frac{53.3 grams of C_{4} H_{10}x416 grams of O_{2} }{116 grams of C_{4} H_{10}}[/tex]

mass of O₂= 191.14 grams

But 191.14 grams of O₂ are not available, 103.9 grams are available. Since you have less mass than you need to react with 53.3 grams of C₄H₁₀, oxygen O₂ will be the limiting reagent.

Mass of carbon dioxide

Then the following rule of three can be applied: if by reaction stoichiometry 416 grams of O₂ form 352 grams of CO₂, 103.9 grams of O₂ form how much mass of CO₂?

[tex]mass of CO_{2} =\frac{103.9 grams of O_{2}x352 grams of CO_{2} }{416grams of O_{2} }[/tex]

mass of CO₂= 87.91 grams

Then, 87.91 grams of CO₂ are produced when 53.3 g of butane is burned in 103.9 g of oxygen.

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