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

Question

Cameron Cochran

Chemistry

8 December, 19:50

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

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Answers (1)

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Francis Escobar · Answer 1

Answer is: mass of burned butane is 11.6 g.

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

m (butane) = 50,0 g.

V (CO ₂) = 17,9 L.

n (CO ₂) = V (CO₂) : Vm.

n (CO ₂) = 17,9 L : 22,4 L/mol.

n (CO ₂) = 0,8 mol.

From chemical reaction n (CO ₂) : n (C₄H₁₀) = 8 : 2.

n (C ₄H₁₀) = 0,8 mol : 4.

n (C ₄H₁₀) = 0,2 mol.

m (C ₄H₁₀) = n (C₄H₁₀) · M (C₄H₁₀).

m (C ₄H₁₀) = 0,2 mol · 58 g/mol.

m (C ₄H₁₀) = 11,6 g.