moles NaNO2 = grams/molar mass?
I wonder why you used 24.627 for the molar volume?
In the Formula...
Na+ + NO2- + HSO3NH2 -> HSO4- + H2O + N2 + Na+
which I know has a 1:1 ratio, How do I figure out the Moles of N2 gas expected.
Molar volume = 24.627 L/mol
Moles of N2 produced = .0067 mol
Volume of N2 gas of STP = .165 L
Original used .5062g of NaNO2 to produce reaction.
Please Help!
5 answers
In the lab experiment (Ideal Gas Behaviour) we were told to calculate the Molar voume of N2 gas using the equation - vol N2 @STP/ mol N2 produced.
The mol of N2 produced were found using the equation PV=nRT
where P = Net pressure of N2 alone = 744.11 (tot pressure - vapour pressure of H2O)
V = .165 L (vol N2 @STP)
R = .082 constant
T = absolute temp of N2 = 295.35
Vol N2 @ STP was found using P1V1/T1 = P2V2/T2
The mol of N2 produced were found using the equation PV=nRT
where P = Net pressure of N2 alone = 744.11 (tot pressure - vapour pressure of H2O)
V = .165 L (vol N2 @STP)
R = .082 constant
T = absolute temp of N2 = 295.35
Vol N2 @ STP was found using P1V1/T1 = P2V2/T2
Thanks. So the 24.627 is your experimentally determined value.
Ok, so how then do I figure out the expected moles for N2 at the end of the equation. Is it that because of the 1:1 ratio that the .0073 mol of NaNO2 are the same for the N2 gas?
That's what I would do.
2 answers