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) :

$\tt \dfrac{0.85}{58,12}=0.015$

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

$\tt \dfrac{8}{4}\times 0.015=0.03$

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

$\tt 0.03\times 22.4~L=0.672~L$