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    1. Convert 100,000 tons/year to kmol/hr

    2. Convert 0.45$/gallon to $/kmol

    Both questions are with regards to methanol. Molar mass = 32.04, density = 791 kg/m3

    I kind of think I know what to do, but I keep getting weird results. I'll post what I've done in a sec, but can someone try it out as well and see what you get?

    My working:

    1. 100,000 \frac {ton}{year} \times 1000 \frac {kg}{ton} = 100,000,000 \frac {kg}{year}
    100,000,000 \frac {kg}{year} \times \frac {year}{365 days} \times \frac {day}{24 hr} = 11,415 \frac {kg}{hr}
    \frac {11,415 \frac {kg}{hr}}{32.04 \frac {kg}{kmol}} = 356 \frac {kmol}{hr}


    2. Let D = $ as latex wont let me use $
     0.45 \frac {D}{gallon} \times 264 \frac {gallon}{m^3} = 78.408 \frac {D}{m^3}
    For some reason I don't know what to do when the $ side stays. But if I take it away and add it back later then I do:
    78.408 m^3 \times 791 \frac {kg}{m^3} = 62,021 \frac {D}{kg}
    \frac {62,021 \frac {D}{kg}}{32.04\frac {kg}{kmol}} = 1937 \frac {D}{kmol}

    Are those reasonable numbers?
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    (Original post by G A B R I E L)
    1. Convert 100,000 tons/year to kmol/hr

    2. Convert 0.45$/gallon to $/kmol

    Both questions are with regards to methanol. Molar mass = 32.04, density = 791 kg/m3

    I kind of think I know what to do, but I keep getting weird results. I'll post what I've done in a sec, but can someone try it out as well and see what you get?

    My working:

    1. 100,000 \frac {ton}{year} \times 1000 \frac {kg}{ton} = 100,000,000 \frac {kg}{year}
    100,000,000 \frac {kg}{year} \times \frac {year}{365 days} \times \frac {day}{24 hr} = 11,415 \frac {kg}{hr}
    \frac {11,415 \frac {kg}{hr}}{32.04 \frac {kg}{kmol}} = 356 \frac {kmol}{hr}


    2. Let D = $ as latex wont let me use $
     0.45 \frac {D}{gallon} \times 264 \frac {gallon}{m^3} = 78.408 \frac {D}{m^3}
    For some reason I don't know what to do when the $ side stays. But if I take it away and add it back later then I do:
    78.408 m^3 \times 791 \frac {kg}{m^3} = 62,021 \frac {D}{kg}
    \frac {62,021 \frac {D}{kg}}{32.04\frac {kg}{kmol}} = 1937 \frac {D}{kmol}

    Are those reasonable numbers?
    Doing it in my head so doing a bit of rounding-

    (1000/32) = 31moles/kg so 31000 moles/tonne so 3.1 x 109 moles in 100000 tonnes

    24 hrs in a day, 365 days/yr so 8760 hrs/yr (approx 9000)

    moles/hr = 3 x 109/9 x 103 = 3.33 x x 105 so 3.33 x x 102 kmol/hr


    The US gallon is about 3.8 litres (I assume it is this because the price is in US$)

    3.8 litres methanol will weigh 3.04 kg (density rounded to 0.8 kg/l)

    3.04 kg methanol is approx 93 moles (from above) so $ 0.45/gallon approx = $0.005/mole therefore $5/kmole

    Hope this is some use
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    (Original post by Cora Lindsay)
    Doing it in my head so doing a bit of rounding-

    (1000/32) = 31moles/kg so 31000 moles/tonne so 3.1 x 109 moles in 100000 tonnes

    24 hrs in a day, 365 days/yr so 8760 hrs/yr (approx 9000)

    moles/hr = 3 x 109/9 x 103 = 3.33 x x 105 so 3.33 x x 102 kmol/hr


    The US gallon is about 3.8 litres (I assume it is this because the price is in US$)

    3.8 litres methanol will weigh 3.04 kg (density rounded to 0.8 kg/l)

    3.04 kg methanol is approx 93 moles (from above) so $ 0.45/gallon approx = $0.005/mole therefore $5/kmole

    Hope this is some use
    We get the same thing for 1 and for 2 yours seems more reasonable. I got that the first time I did it, but it just thought that a kmole of something would cost more than $5.

    Could you help with one more thing?

    How would you convert 0.15 $/lb in $/gallon. 791 kg/m3 = 6.6 lb/gallon so i get basically 1$/gallon
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    (Original post by G A B R I E L)
    How would you convert 0.15 $/lb in $/gallon. 791 kg/m3 = 6.6 lb/gallon so i get basically 1$/gallon
    Sounds about right

    C
 
 
 
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