A 3.0 mol sample of KCIO3 was decomposed according to the equation

N

Nikifor

New member
You have a typo in the question.
anyways.

We create a proportion

3mol/2 (KClO3)= x/3(O2)

x=4.5 mol.

Answer: n(O2)= 4.5 mol.
 
2KCIO3(s)=2KCI(s)+3)2(g) how many moles of? O2 are formed assuming 100% yield?
Sorry it was 3O2(g)..thank you could you please show me how you came up with that answer..i have a test tomorrow..thanks
 
You have a typo in the question.
anyways.

We create a proportion

3mol/2 (KClO3)= x/3(O2)

x=4.5 mol.

Answer: n(O2)= 4.5 mol.
 
You have a typo in the question.
anyways.

We create a proportion

3mol/2 (KClO3)= x/3(O2)

x=4.5 mol.

Answer: n(O2)= 4.5 mol.
 
You have a typo in the question.
anyways.

We create a proportion

3mol/2 (KClO3)= x/3(O2)

x=4.5 mol.

Answer: n(O2)= 4.5 mol.
 
You have a typo in the question.
anyways.

We create a proportion

3mol/2 (KClO3)= x/3(O2)

x=4.5 mol.

Answer: n(O2)= 4.5 mol.
 
You have a typo in the question.
anyways.

We create a proportion

3mol/2 (KClO3)= x/3(O2)

x=4.5 mol.

Answer: n(O2)= 4.5 mol.
 
You have a typo in the question.
anyways.

We create a proportion

3mol/2 (KClO3)= x/3(O2)

x=4.5 mol.

Answer: n(O2)= 4.5 mol.
 
You have a typo in the question.
anyways.

We create a proportion

3mol/2 (KClO3)= x/3(O2)

x=4.5 mol.

Answer: n(O2)= 4.5 mol.
 
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