Question

Volume in liters of carbon dioxide produced at stp when 0.85 grams of butane reacts

Answer 1

Volume in liters of Carbon dioxide : 0.672

Further explanation

Reaction(combustion of butane-C₄H₁₀)

2C₄H₁₀+13O₂⇒8CO₂+10H₂O

mol butane (MW=58,12 g/mol) :

`\tt (0.85)/(58,12)=0.015`

mol CO₂ : mol C₄H₁₀ = 8 : 4, so mol CO₂ :

`\tt (8)/(4)* 0.015=0.03`

At STP, 1 mol = 22.4 L, so volume CO₂ :

`\tt 0.03* 22.4~L=0.672~L`