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16 December, 23:39

Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is 2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (l) At 1.00 atm and 23 oC, what is the volume of carbon dioxide formed by the combustion of 3.4 g of butane?

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  1. 17 December, 00:25
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    5.638 L is the volume formed

    Explanation:

    The reaction for the combustion is:

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

    First of all we convert the mass of butane to moles (mass / molar mass)

    3.4 g / 58 g/mol = 0.058 moles

    As we assume, the oxygen in excess ratio between butane and carbon dioxide is 2:8. Let's make a rule of three:

    2 moles of butane can produce 8 moles of carbon dioxide

    0.058 moles of butane must produce (0.058. 8) / 2 = 0.234 moles of CO₂

    Now we apply the Ideal Gases Law to find out the volume formed.

    P. V = n. R. T

    1atm. V = 0.234 mol. 0.082 L. atm/mol. K. 296K

    V = 0.234 mol. 0.082 L. atm/mol. K. 296K / 1atm = 5.68 L
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