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Chemistry Question (Gaw Laws?) - What volume of water vapour at satp conditions could be made from...?

The density of butance is 2.489 mg/ml at 15.0 degrees C. What volume of water vapour at SATP conditions could be made from 1.893 mL of butance at 15.0 degrees C?

I feel like I'm given.... dare I say... too much information.

I feel like this is a gas law question but not sure. Help?? Assignment due tomorrow.

Detailed steps would be perfect, please!

Thanks in advance :)

3 Answers

  • Lexi R
    Lv 7
    9 years ago
    Favorite Answer

    You don't really need the density to do this one, though you can use the density to calculate the mass and then moles of butane and work the problem that way, but there is a simpler way to do it...

    first write a balanced equation for the reaction

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

    The same volume of any ideal gas under the same conditions of temp and pressure has the same number of moles. So, as long as temp and pressure are constant volumes of gases react in the same ratios as moles, so you can apply the molar ratios the gases react in directly to the volumes.

    2 moles butane react to produce 10 moles H2O

    So volume H2O = 10/2 x volume butane

    = 10/2 x 1.893 ml

    = 9.465 ml

    Thus, at 15 deg C you will produce 9.465 ml of H2O

    Now use V1 / T1 = V2 / T2

    to calculate the volume when the temp is increase to SATP conditions (you are not told the pressure at 15 deg C so we need to assume it is constant (pressure at SATP = 1 atm)

    V1 = 9.465 ml

    T1 = 15 deg C = 288.15 K

    V2 = ? ml

    T2 = 25 deg C = 298.15 K (temperature at SATP (standard ambient temperature and pressure)

    V2 = V1T2 / T1

    = 9.465 ml x 298.15 K / 288.15 K

    = 9.79 ml

  • 9 years ago

    I agree that you probably can do without knowing the density of the butane. This information has been given to you if you chose to find the mass of butane in 1.893 mL of it at 15.0°C. However, in this case it is not necessary to do.

    First, write the balanced equation for the combustion of butane:

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

    When dealing with gases, the moles of gases are proportional to volumes of gases. We can find the volume of H2O produced from 1.893 mL of C4H10 by using a simple proportion:

    (1.893 mL C4H10) : (X mL H2O) = (2 mol C4H10) : (10 mol H2O)

    X mL H2O = {(1.893 mL C4H10)(10 mol H2O)} / (2 mol C4H10)

    X mL H2O = 9.465 mL H2O at 15.0°C

    Assuming that the reaction happened at constant pressure, we can use Charles's Law equation to calculate the volume of water vapor at STP:

    (V1) / (T1) = (V2) / (T2)

    At laboratory conditions:

    V1 = 9.465 mL

    T1 = 15.0°C + 273 = 286 K

    AT STP conditions:

    V2 = ?

    T2 = 273 K

    V2 = {(V1)(T2)} / (T1)

    V2 = {(9.465 mL)(273 K)} / (286 K)

    V2 = 9.0347727 mL or 9.03 mL rounded to three significant figures

    Answer: About 9.03 mL of water vapor at STP conditions could be made from 1.893 mL of butane at 15.0°C.

    Source(s): personal knowledge
  • 4 years ago

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