Chem
posted by Sara .
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%

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). 
First, start with a balanced equation
2KClO_{3}(s)> 2KCl(s)+ 3O_{2}(g)
2KClO_{3}: mm=2(39+35.5+16*3)= 245
3O_{2}: mm = 3(2*16) = 96
Theoretical yield
=2.14*(96/245)
=0.839
Actual yield
=0.67
Percent yield
=0.67/0.839
=80% 
80.% (to two significant figures)