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27 July, 17:39

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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  1. 27 July, 21:17
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    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.
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