posted by SadFace .
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 = ??
I guess that's good?
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