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    andreinhamtb's Avatar
    andreinhamtb Posts: 1, Reputation: 1
    New Member
    		  
    #1

    Feb 26, 2010, 08:43 AM
    stoichiometry questions
    I have a stoichiometry problem to solve...
    I need to find a b and c and x

    a KCLOx
    HTML Code:
    reacts and form
    --- bKCL + cO2

    x could be 1, 2, 3, or 4.
    O2 is oxygen gas
    Mass of the sample (KCLOx) = 1.0318g
    mass of Kcl produced = 0.6251g
    Mass of oxygen produced = 0.4067g

    I found that
    moles of KCL produced is = 0.008385g
    moles of O2 produced is = 0.01459g
    but I can't find the ratio of moles O2/moles KCL

    Thanks
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
    Uber Member
    		  
    #2

    Feb 27, 2010, 01:51 AM

    First, the symbol for chlorine is Cl, not CL.

    Moles are measured in moles, not in grams.

    I get: moles of KCl = 0.6251/(39.1+35.5) = 8.379 x 10^-3mol
    moles of O2 = 0.4067/32 = 0.01271 mol
    moles of KClOx = 1.0318/(39.1+35.5+ (x*16)) mol

    Mole ratio of KCl to O2 = 8.379/0.01271 = 0.659 = 2:3

    This gives:

    a KClOx -> 2 KCl + 3 O2

    a must be 2

    2 KClOx -> 2 KCl + 3 O2

    So, O must be 3.

    ClO3 is the chlorate (V) ion, which exists.
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