普通化學問題? 我用英文問你用中文回答20點

Isooctane, C8H18, is the component of gasoline from which the term

octane rating derives.

(a)write a balanced equation for the combustion of isooctane to yield

CO2 <----二氧化碳 and H2O <---水.

(b)assuming that gasoline is 100% isooctane , that isooctane burns to

produce only CO2 and H2O, and that the density of isooctane is

0.792 g/ml, what mass of CO2(in kilograms) is produced each year by

the annual U.S. gasoline consumption of 4.6*10的10次方L ?

(c)what is the volume (in liters) of this CO2 at STP ?

(d)how many moles of air are necessary for the combustion of 1 mol

of isooctane, assuming that air is 21.0% O2 by volume ? what is the

volume (in liters) of this air at STP ?

拜託幫幫忙...快考試了....

1 Answer

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  • ?
    Lv 4
    1 decade ago
    Best Answer

    a)C8H18 + 12.5 O2  8 CO2 + 9 H2O

    or, C8H18 + 12.5 O2  8 CO2 + 9 H2O

    2007-12-07 12:48:14 補充:

    b) 10

    (4.6X10 L)X(1000mL/LX0.792g/mL)X(1molC8H18/116gC8H18)X(8molCO2/1molC8H18)X(44gCO2/1molCO2)x(1kgCO2/1000gCO2)

    11

    =1.11x10 kg

    2007-12-07 12:49:27 補充:

    1.11x10的11次方kg (這裡沒辦法貼的完整)

    2007-12-07 12:51:58 補充:

    c)

    pV = nRT

    here, p = 1 atm

    n = 1.11 x 1011 kg CO2 = 2.51 x 1012 mol CO2

    T = 25oC = 298 K

    V=nRT/p =6.14x10的13次方L

    2007-12-07 12:57:00 補充:

    D)

    因為12.5 moles O2 燃燒生成 8 moles CO2

    O2 V = 12.5/8 x 6.14 x 10的13 次方 L = 9.59 x 10的13 次方 L.

    空氣中含氧量 21.0% ,

    所以需要空氣 V=9.59 x 10的13 次方L / 0.21 = 4.57 x 10的14次方 L.

    2007-12-07 12:57:56 補充:

    希望我沒算錯 祝福妳

    Source(s): , 20年前的大學普化知識
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