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SpiderQuack · Sep 18, 2010, 08:13 PM

How many moles of O2 were produced in a decomposition reaction of H2O2 if the barometric pressure was 0.980 atm, the temperature was 298 K and the volume of O2 gas collected was 0.0500 L?

Unknown008 · Sep 19, 2010, 06:16 AM

Use the ideal gas equation.

$PV = nRT$

R = 8.31
You have the pressure, the volume and the temperature. Find n, the number of moles. Post your answer! :)

karely111 · Jan 13, 2011, 11:23 AM

so it's be n= PV/RT and if you plug them in it'd be pressure= P =0.980 atm times volume= V = 0.0500L which would give you 0.049 and then divide by the multiplication of 8.31 times temperature= T = 298 Kelvins which would be 2476.38, right? So ((0.980)(0.05))/((8.31)(298))= .000019787 = 2 x 10^-5... Okay I just did it in order to see if I could get it right, but it doesn't look right to me, uhm so please could someone just tell me if I'm wrong or right? Thank you so much

Unknown008 · Jan 13, 2011, 11:28 AM

No, it's not right because you didn't take into consideration the units.

The value of R is 8.31 when taking pressure in Pascals, volume in m^3, n in moles and T in Kelvin.

Can you try again?

1 atm = 101300 Pa

karely111 · Jan 13, 2011, 11:37 AM

Okay, so I'd have to change 0.980 atm into Pascals? Which would be 99274 Pa?
And I would also have to change the volume 0.05 L to m^3?

Unknown008 · Jan 13, 2011, 11:40 AM

Yes, now

1 L = 0.001 m^3

karely111 · Jan 13, 2011, 11:50 AM

So it'd be 50m^3, right? I plug it in and I get 2004.42 so I believe I'm doing something wrong again. T_T

Unknown008 · Jan 13, 2011, 11:55 AM

You got it the wrong way round ;)

$1\ L \to 0.001\ m^3$

$0.05\ L \to 0.001 \times 0.05 = 5\times 10^{-5}\ m^3$

karely111 · Jan 13, 2011, 11:59 AM

OH! I got that and I thought I was wrong! Thank you so much now let me see...

karely111 · Jan 13, 2011, 12:00 PM

Nope uhm... 0.002?

Unknown008 · Jan 13, 2011, 12:01 PM

Yes, that's what I got :)

Don't forget the units, 0.002 moles