hope this helped.:)
Step-by-step explanation:
Yes, the number of moles of oxygen gas produced by your reaction under those conditions for pressure and temperature will be 0.0025.
Hydrogen peroxide,
H
2
O
2
, decomposes to give water and oxygen gas according to the balanced chemical equation
2
H
2
O
2
(
a
q
)
→
2
H
2
O
(
l
)
+
O
2
(
g
)
You've collected 0.061 L of oxygen gas at 295.15 K and 1 atm, so you've got all the data you need to calculate the number of moles of oxygen gas produced by using the ideal gas law equation
P
V
=
n
R
T
⇒
n
=
P
V
R
T
n
O
2
=
1
atm
⋅
0.061
L
0.082
L
⋅
atm
mol
⋅
K
=
0.0025 moles
So, if this was your first question, then yes, your reaction produced 0.0025 moles of oxygen gas.
I find the second part of your question to be a little confusing. You were given the density of the hydrogen peroxide solution, so are you supposed to use that to determine the theoretical number of moles of oxygen for this reaction?
I'm not sure what
100%
H
2
O
2
=
1.02 g/mL
means, do you have a certain volume of hydrogen peroxide solution?
SIDE NOTE According to the additional information posted by Heather, it turns out that the initial hydrogen peroxide solution had a volume of 5 mL.
Even with the volume of the initial solution, you'd need its percent concentration to try and determine exactly how many moles you had present before the reaction.
Once you know how many moles of hydrogen peroxide you had, assume that all of the react and use the
2
:
1
mole ratio that exists between
H
2
O
2
and
O
2
to get the number of moles of oxygen your reaction could have produced.