Suppose a grill lighter contains 50.0 g of butane. how many grams of butane in the lighter would have to be burned to produce 17.9 l of carbon dioxide at stp

Answer is: mass of burned butane is 11.6 g.

Chemical reaction: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O.

m(butane) = 50,0 g.

V(CO₂) = 17,9 L.
n(CO₂) = V(CO₂) ÷ Vm.
n(CO₂) = 17,9 L ÷ 22,4 L/mol.
n(CO₂) = 0,8 mol.
From chemical reaction n(CO₂) : n(C₄H₁₀) = 8 : 2.
n(C₄H₁₀) = 0,8 mol ÷ 4.
n(C₄H₁₀) = 0,2 mol.
m(C₄H₁₀) = n(C₄H₁₀) · M(C₄H₁₀).
m(C₄H₁₀) = 0,2 mol · 58 g/mol.
m(C₄H₁₀) = 11,6 g.

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