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Chem

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Determine the percent yeild of:
KCl3(s)--> KCl(s)+ O2(g)
2.14g of KCl3 produces 0.67 of O2

I get 123% when the correct answer is 80%

My work:
KCl3: mm = 122.45, n =0.017

Theoretical yield: .017mol * 32g/mol

Pecent Yield: 0.67g/ 0.544g *100% = 123%

  • Chem - ,

    Thanks for showing your work. It helped me spot the error right off. You didn't balance the equation. Also note that I corrected the formula for KClO3 but you have the right moles so you must have used the correct molar mass.
    2KClO3 ==> 2KCl + 3O2

    moles KClO3 = 2.14/122.5 = 0.0175

    moles oxygen = 0.0175 x (3 moles O2/2 moles KClO3) = 0.0175 x (3/2) = 0.0262.

    grams O2 = moles O2 x molar mass = 0.0262 x 32 = 0.838 g O2 for theoretical yield.
    Use that number with 0.67 and see if it's ok. I get 79.9% which rounds to 80% to two significant figures (which is all we are allowed with the 0.67).

  • Chem - ,

    First, start with a balanced equation
    2KClO3(s)--> 2KCl(s)+ 3O2(g)

    2KClO3: mm=2(39+35.5+16*3)= 245
    3O2: mm = 3(2*16) = 96

    Theoretical yield
    =2.14*(96/245)
    =0.839
    Actual yield
    =0.67
    Percent yield
    =0.67/0.839
    =80%

  • Chem-corr. - ,

    80.% (to two significant figures)

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