2C4H10 (g) + 13O2 (g) → 8CO2 (g) + 10H2O (g) 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?

A) 46.4 g

B) 8.00 g

C) 8.95 g

D) 11.6 g

Answer is: mass of butane is D) 11.6 g.

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.