When 0.5963g of a sample containing NaNO2 and some inert material is allowed to react with excess HSO3NH2, the volume of N2 gas produced is measured. The volume of N2 gas corrected to 298.15K and 100kPa is calculated to be 0.146L. Find the mass percent of NaNO2 in the sample using the molar volume of N2 at 298.15K and 100kPa (from the previous question). The molar volume from the previous question was 22.8 L/mol.
It would help greatly if you had an equation for the reaction.
NaNO2 + HSO3NH2 = NaHSO4 + H2O + N2
How many moles N2 do you have? That is 0.146L x (1 mol/22.8L) = ?? moles N2.
Using the coefficients in the balanced equation, convert moles N2 to moles NaNO2. (Hint: that is 1:1 ratio).
Convert moles NaNO2 to grams. grams = moles x molar mass.
%NaNO2 = (grams NaNO2/mass sample)*100 = ??
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To find the mass percent of NaNO2 in the sample, we need to use the information given about the reaction and the volume of N2 gas produced.
First, let's determine the moles of N2 gas produced using the ideal gas law:
PV = nRT
Where:
P = pressure (100 kPa)
V = volume of N2 gas corrected to standard conditions (0.146 L)
n = moles of N2 gas (unknown)
R = ideal gas constant (0.0821 L·atm·K^–1·mol^–1)
T = temperature (298.15 K)
Rearranging the equation to solve for moles (n):
n = PV / RT
n = (100 kPa * 0.146 L) / (0.0821 L·atm·K^–1·mol^–1 * 298.15 K)
Now, let's calculate the moles of N2 gas produced.
n = (14.6 kPa·L) / (0.0821 L·atm·K^–1·mol^–1 * 298.15 K)
n ≈ 0.5805 mol
Since the balanced chemical equation for the reaction is not given, we'll assume it follows this stoichiometry:
2NaNO2 + 2HSO3NH2 → N2 + 2H2O + 2NaHSO4
From the stoichiometry, we can see that for every mole of N2 produced, there are 2 moles of NaNO2 in the sample.
Therefore, the moles of NaNO2 in the sample would be half the moles of N2 gas produced:
moles of NaNO2 = 0.5805 mol / 2
moles of NaNO2 ≈ 0.2902 mol
Now, to find the mass of NaNO2, we'll use the molar mass of NaNO2 which is approximately 69 grams/mol.
mass of NaNO2 = moles of NaNO2 * molar mass of NaNO2
mass of NaNO2 ≈ 0.2902 mol * 69 g/mol
Finally, to find the mass percent of NaNO2 in the sample, we'll divide the mass of NaNO2 by the total mass of the sample and multiply by 100.
mass percent of NaNO2 = (mass of NaNO2 / total mass of the sample) * 100
mass percent of NaNO2 ≈ (0.2902 mol * 69 g/mol / 0.5963 g) * 100
Therefore, the mass percent of NaNO2 in the sample is approximately 34.5%.