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.
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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 → N2 + ?
NaNO2 is composed of Na +1 + NO2 -1
NO2 -1 is composed of N +3 and 2 O -2
HSO3NH2 is composed of NH2-1 and HSO3+1
NH2 -1 is composed of N -3 and 2 H+1
HSO3 +1 is composed of H+1 + S +6 + 3 O -2
N2 is composed of 2 N 0
N+3 → N2 0
and
N-3 → N2 0
This is a redox reaction
2 NO2-1→ N2 + 4 H2O
8 H+1 + 2 NO2-1→ N2 + 4 H2O
6 e- + 8 H+1 + 2 NO2-1→ N2 + 4 H2O
and
2 NH2-1 → N2
2 NH2-1 → N2 + 4 H+1
2 NH2-1 → N2 + 4 H+1 + 6 e-
Add both equations
8 H+1 + 2 NO2-1 + 2 NH2-1 → N2 + 4 H2O + N2 + 4 H+1
4 H+1 + 2 NO2-1 + 2 NH2-1 → 2 N2 + 4 H2O
4 moles of H+1 + 2 moles of NO2-1 + 2 moles of NH2-1 react to produce 2 moles of N2 + 4 moles of H2O
The other ions are spectator ions.
1 mole of N2 came from 2 moles of NO2-1
and
1 mole of N2 came from 2 moles of NH2-1
1 mole of NaNO2 contains 1 mole of Na+1 and1 mole of NO2-1!
Molar mass of NaNO2 = 23 + 14 + 32 = 69 g
Moles of NaNO2 = 0.5693 ÷ 69
So moles of NO2 = 0.5693 ÷ 69
So moles of N2= ½ * (0.5693 ÷ 69)
and
Moles of NH2 -1 = moles of NO2 -1
(0.5693 ÷ 69) moles of NH2 -1 will produce ½ * (0.5693 ÷ 69) moles of N2
Total moles of N2 = (0.5693 ÷ 69)
Use gas law equations to determine Volume
Moles of HSO3NH2 which reacted = moles of NaNO2 which reacted = (0.5693 ÷ 69)
Determine the mass of (0.5693 ÷ 69) moles of HSO3NH2
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.
NaNO2 + HSO3NH2 → N2 + ?
NaNO2 is composed of Na +1 + NO2 -1
NO2 -1 is composed of N +3 and 2 O -2
HSO3NH2 is composed of NH2-1 and HSO3+1
NH2 -1 is composed of N -3 and 2 H+1
HSO3 +1 is composed of H+1 + S +6 + 3 O -2
N2 is composed of 2 N 0
N+3 → N2 0
and
N-3 → N2 0
This is a redox reaction
2 NO2-1→ N2 + 4 H2O
8 H+1 + 2 NO2-1→ N2 + 4 H2O
6 e- + 8 H+1 + 2 NO2-1→ N2 + 4 H2O
and
2 NH2-1 → N2
2 NH2-1 → N2 + 4 H+1
2 NH2-1 → N2 + 4 H+1 + 6 e-
Add both equations
8 H+1 + 2 NO2-1 + 2 NH2-1 → N2 + 4 H2O + N2 + 4 H+1
4 H+1 + 2 NO2-1 + 2 NH2-1 → 2 N2 + 4 H2O
4 moles of H+1 + 2 moles of NO2-1 + 2 moles of NH2-1 react to produce 2 moles of N2 + 4 moles of H2O
The other ions are spectator ions.
1 mole of N2 came from 2 moles of NO2-1
and
1 mole of N2 came from 2 moles of NH2-1
1 mole of NaNO2 contains 1 mole of Na+1 and1 mole of NO2-1!
Molar mass of NaNO2 = 23 + 14 + 32 = 69 g
Moles of NaNO2 = 0.5693 ÷ 69
So moles of NO2 = 0.5693 ÷ 69
So moles of N2= ½ * (0.5693 ÷ 69)
and
Moles of NH2 -1 = moles of NO2 -1
(0.5693 ÷ 69) moles of NH2 -1 will produce ½ * (0.5693 ÷ 69) moles of N2
Total moles of N2 = (0.5693 ÷ 69)
Use gas law equations to determine Volume
Moles of HSO3NH2 which reacted = moles of NaNO2 which reacted = (0.5693 ÷ 69)
Determine the mass of (0.5693 ÷ 69) moles of HSO3NH2
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.