Easy Chemistry Question

[ironizer]

If a KClO3/KCl mixture weighing 1.64 g loses 0.315 g upon heating, what is the % KClO3 in the mixture?

 

I can't figure out what this means. What is it even asking?

[hermanntrude]

KClO3 decomposes on heating to give KCl and O2:

 

[ce]KClO3 -> KCl + 3/2O2[/ce]

 

If this doesnt help, let me know where the trouble is and i'll try to help some more.

[ironizer]

I don't even see what the question is asking. What does it mean, "what's the % KClO3 in the mixture?" ? Is it KClO3 mixed with KCl or what?

[UC]

Yes, you have a mixture of KCl and KClO3. When heated, KClO3 decomposes to KCl and oxygen by the equation hermanntrude posted. The only thing that is going to leave the mixture is oxygen gas, so it means that 0.315g of oxygen is formed. What percent of the weight of the original mixture was KClO3?

[vedmecum]

Ans. is 49.02 percent . From the equation 1 mole i.e. 122.5gm of KClO3 produces 1.5 mole of oxygen i.e. 48 gm . So 0.315 gm of oxygen will be produced if 0.804 gm of KClO3 will be COMPLETELY burned ( apply unitary method 0.315*122.5/48=0.804) . So the percentage is 0.804*100/1.64=49.02.

[insane_alien]

vedmecum, when someone is asking a homework question you do not just throw them the answer. point them in the right direction and throw out a few hints sure, but do not do their work for them.

 

we go by the old adage 'give a man a fish and he will eat for a day, teach a man to fish and he will never go hungry again' sort of thing.

[ironizer]

Ye true, UC's answer is what I was looking for. Now it makes sense.

 

But the answer is still useful, I don't really need to learn to "fish" here because chemistry is BS and as soon as I'm done taking the required 2 quarters of chem I'll kick chemistry in the nuts and ditch it for good. Then I can go do my physics/maths and all the other useful studies.

 

49.0 was what I had to enter in, because they give the grams to 3 significant figures. Now I'm done and I can go watch some of the Australian Open. Go Federer.