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.
NaNO2 + HSO3NH2 = NaHSO4 + H2O + N2