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How much Chromium metal will produced from the above reaction?

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Chromium metal, Cr(s) can be prepared from reacting chromium oxide, Cr2O3(s), with aluminum, Al(s). Al2O3(s) is a by-product. Assume a reaction of 20,0 kg of chromium oxide with 5,00 kg of aluminum metal. 1)how much Chromium metal will produced from the above reaction? 2) who reagent is too much and how much is it?


    2. Relevant equations
    please help me im confused is the second solution right ?

    3. The attempt at a solution
    one thought is that

    Cr2O3 + 2 Al -> 2 Cr + Al2O3

    M1= M(Cr2O3)= 152 g/mol
    M2= M(Al)= 27 g/mol
    -> n1 = 20E3/152 = 131.58 mol
    -> n2= 5000/27 = 185.18 mol= number of mols Cr formed
    mass of cr formed:m3=n2*M3 = 185.18 mol *52 g/mol= 9629 g

    Now, according to stoichiometry n1/n2 = 1/2 ; n2=185.18 mol -> n1 = 92.59 mol (m1=n1*M1= 12222 g) so 7778 g Cr2O3 are in excess.

    and one other is


    Cr2O3(s) + 2 Al(s) = 2 Cr(s) + Al2O3(s)

    2 Cr = 2 (52.00) = 104.00
    3 O = 3 (16.00) = 48.00
    ------
    152.00

    1000 g 20,0 kg Cr2O3(s) = 20,0 kg * (1000 kg / 1g) = 20.000g Cr2O3(s)
    -----------
    1 kg
    5,00 kg Al(s) = 5,00 kg * (1000 kg / 1g) = 5.000g Al


    20.000 g Cr2O3(s) * 1 mol Cr2O3(s) / 152 g Cr2O3(s)* 2 mol Cr(s) / 1 mol Cr2O3(s) = 0.26314 mol Cr(s)

    5.000 g Al(s) * 1 mol Al / 27 g Al(s) * 2 mol Cr / 2 mol Al(s) = 0.1851 mol Cr(s)

    Cr2O3(s) is reactant excess

    0.1851 mol Cr(s) * 1 mol Cr2O3(s) / 2 mol Cr(s) * 152g Cr2O3(s) / 1 mol Cr2O3(s) = 14.06 g

    20.00 g - 14.06 g = 5.94 g Cr2O3(s) is how much excess left

    0.1851 mol Cr * 52 g Cr(s) / 1 mol Cr = 9.6252 g Cr(s) was formed
     
  2. jcsd
  3. Nov 2, 2008 #2

    Borek

    User Avatar

    Staff: Mentor

    Beware when using dot and comma in numbers, 20.000 is not the same as 20000.
     
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