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Anonymous asked in Science & MathematicsChemistry · 1 decade ago

what volume of oxygen gas at 25C and 1.04 atm is needed for the complete combustion of 5.53 of propane?

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  • Anonymous
    1 decade ago
    Favorite Answer

    Pv=nrT

    P=1.04

    V=unknown

    n=5.53(if this is in moles, if in grams convert to moles knowing propane =C3H8 and use stoichiometry: (5.53gC3H8)*(1 mol/44gramsC3H8)

    r=constant .0821

    T=298K(must be in Kelvin, add degrees celsius+273)

    (1.04)(V)=moles(.0821)(298)

    solve for V

    if moles is 5.53 answer should be 130L

    if moles is .1257 answer should be 2.96L

    hope this helps!

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