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9. What volume of oxygen (in cm3) is required for the complete combustion of 2.32 g of butane?

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

Molar mass of butane = (4 x 12) + (10 x 1) = 58 g/mol

Moles = mass / molar mass = 2.32 / 58 = 0.04 mol

Reacting ratio of C4H10 to O2 = 2 : 13 = 1 : 6.5

Moles of O2 = 0.04 x 6.5 = 0.26 mol

Volume = moles x 24 = 0.26 x 24 = 6.24 cm3

6 Answers

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  • Anonymous
    1 month ago

    Thought this was a botox question so I will ask the dog as I have no clue

  • Mike A
    Lv 7
    1 month ago

    You were doing OK up to the last line where the figure of 24 comes out of nowhere with no units. One mole of an ideal gas will occupy 22.4 Litres at STP. That's 22400 CM3 and 0.26 mol will occupy 5.82 Litres at STP. The question should state at what conditions the volume of O2 is measured and you need to convert using PV = nRT the ideal gas law.

  • Dr W
    Lv 7
    2 months ago

    balanced reaction

    .. 2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O 

    then

    .. 2.32g C4H10... 1 mol C4H10.. .. 13 mol O2

     ----- ---- ---- ---- x ----- ---- ---- ---- x ---- ---- ---- -- = 0.2595 mol O2

    .. .... .. .1.. ... .. .. .58.12g C4H10.. .2 mol C4H10

    you can't convert that to cm3 of O2 without more information

    pressure, temperature, for example

  • 2 months ago

    (2.32 g C4H10) / (58.1222 g C4H10/mol) x (13 mol O2 / 2 mol C4H10) =

    0.25945 mol O2

    You do not specify the conditions under which the volume of O2 is measured.  So supposing 1 atm and 0°C:

    (0.25945 mol O2) x (22.414 L/mol) = 5.8153 L = 5.82 L O2

    So you are correct up to the last line.  It's not clear to me where "24" actually came from, but it's a reasonable value for liters per mole, using slightly different conditions than I assumed.  But the thing that makes you answer 1000 times wrong, is the unit "cm3".  You have used no units in the calculation, so you are confused about the final answer.  It must be somewhere around 6 LITERS.  You MUST include units in all your calculations.

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  • 2 months ago

    Butane + Oxygen ➜ Carbon Dioxide + water

    2C₄H₁₀ + 13O₂ ➜ 8CO₂ + 10H₂O

    molecular weights

    C = 12

    H = 1

    O = 16

    2C₄H₁₀ = 2•58 = 116

    13O₂ = 13•32 = 416

    8CO₂ = 8•44 = 352

    10H₂O = 10•18 = 180

    check 116+416 = 352+180 = 532

    116 grams of C₄H₁₀ + 416 grams of O₂ ➜ 352 grams of CO₂ + 180 grams of H₂O

    2 mole of C₄H₁₀ + 13 moles of O₂ ➜ 8 mole of CO₂ + 10 moles of H₂O

    gram ratio of oxygen to butane is 416/116

    416/116 = x/2.32

    x = 8.32 g

    but to get volume, you need temperature and pressure.

    IF I assume STP, old version,  (but you should so state)

    8.32 g / 32 g/mol = 0.26 mol

    0.26 mol x 22.41 L/mol = 5.83 L or 5830 mL or 5830 cm³

    One mole of any ideal gas at STP has a volume of 22.41L  (old def of STP, 1 atm)STP = 0ºC & 1 ATM   (new 100 kPa)

  • 2 months ago

    Yes is correct.

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