[LoraSaur]

I am having an issue with a chemistry problem. I hope someone can *delete me*  I have tried to work this problem from a few different angles but I get halfway through and assume I am doing something terribly wrong.


What mass (in grams) of KClO3 (molecular weight=122.55 g/mole) would be required to decompose to produce 30.5L O2 measured at 950 torr and 295.15K.

The given equation is:

2KClO3(s) = 2KCl(s) + 3O2(g)

I am not sure how to set up the problem correctly. Any help would be greatly appreciated!

[Arkcon]

OK, in order to begin to solve this yourself, can you write out the Gas Law formula?  You know this is a gas law problem, so can you write out the formula, and begin to substitute values you know into the correct places in the formula?  You know what goes where, because the units will match.  There may be some conversions for you to do, but we'll get to that.

[LoraSaur]

I was attempting to use the Ideal Gas Law solving for n with the equation n=PV/RT converting to mass from moles, but that didn't seem to work right.  This is as far as I got:

(30.5L O2)(850 torr/295.15K)(mole x K/0.08206 mole x K)(1 mole O2 /32 g/mole)(2 mole KClO3/3moles O2)(1mole KClO3/ 122.5g/mole)

Am I on the right track?  It seems wrong to me.

[fledarmus]

(30.5L O2)(850 torr/295.15K)(mole x K/0.08206 mole x K) (1 mole O2 /32 g/mole) (2 mole KClO3/3moles O2)(1mole KClO3/ 122.5g/mole)

Check your units where indicated - and one of your conversion factors has a problem, mole KClO₃ isn't canceling the way you have it written. You also still have a "torr" to convert...

[LoraSaur]

Thanks, I knew I was missing something. 

Does this look better? 

(30.5L O2)(850 torr/295.15K)(mole x K/0.08206 L x atm)(2 mole KClO3/3moles O2)(122.5g /1 mole KClO3)(1 atm/760 torr)

[fledarmus]

Much better 

[LoraSaur]

I'll take that as a good sign.   

My answer is 115.0206195

[sjb]

I'll take that as a good sign.   

My answer is 115.0206195

Sig figs? Units (?)

[LoraSaur]

I did 3 sig figs so 115 g KClO3

Thanks for everyone's *delete me*