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!
moles NaNO2 = grams/molar mass?
I wonder why you used 24.627 for the molar volume?
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
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.
To find the moles of N2 gas expected in the given reaction, you need to use stoichiometry. The balanced equation provided shows that 1 mole of NaNO2 reacts to produce 1 mole of N2 gas.
Step 1: Calculate the moles of NaNO2 used
To find the moles of NaNO2, you can use the given mass of NaNO2 and its molar mass. The molar mass of NaNO2 can be calculated by adding up the atomic masses of each element: Na (sodium) has a molar mass of 22.99 g/mol, N (nitrogen) has a molar mass of 14.01 g/mol, and O (oxygen) has a molar mass of 16.00 g/mol.
Molar mass of NaNO2 = 22.99 g/mol (Na) + 14.01 g/mol (N) + (16.00 g/mol (O) × 2) = 69.00 g/mol
Next, you can use the given mass of NaNO2 (0.5062 g) and its molar mass to calculate the moles of NaNO2:
Moles of NaNO2 = Mass of NaNO2 / Molar mass of NaNO2 = 0.5062 g / 69.00 g/mol = 0.00735 mol
Step 2: Determine the moles of N2 gas
Since the balanced equation shows a 1:1 ratio between NaNO2 and N2 gas, the moles of N2 gas produced will be the same as the moles of NaNO2 used. In this case, it is 0.00735 mol.
So, the moles of N2 gas expected in the reaction is 0.00735 mol.
Note: The given value of 0.0067 mol for the moles of N2 produced might be a calculated value, possibly considering the given volume of N2 gas at STP.
I hope this explanation helps!