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?

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! :)

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

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

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?

Yes, now

1 L = 0.001 m^3

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

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

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

Nope uhm... 0.002?

Yes, that's what I got :)

Don't forget the units, 0.002 moles