# 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 -

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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