Homework Answers
1. Moles of HSO3NH2 = Mass/ Molar Mass of the compound = (1.41 g)/ ( 97.1 g / mole) = 0.0145 mole
The reaction is: HSO3NH2 (aq) + NaNO2 (aq) = NaHSO4 (aq) + N2 (g) + H2O (l)
So, moles of NaNO2 reacted = 0.0145 moles
Volume of NaNO2 solution required = Moles / Molarity of the solution = 0.0145 moles/ (0.80 M) = 0.0181 L = 18.1 mL
2. Moles of N2 produced (n) = 0.0145 mole
Temperature = 17.5 0C = (17.5 + 273) K = 290.5 K
Pressure = (Barometric pressure - Vapour pressure at 17.5 0C) = (782.5 - 15) mmHg = 767.5 mmHg = (767.5/ 760) atm = 1.01 atm
Volume (V) = n * R *T / P = (0.0145 mole) * (0.082 Lit atm mole-1 K-1) * ( 290.5 K) / (1.01 atm) = 0.342 L
So, volume of N2 collected = 0.342 Liters = 342 mL
3. Partial pressure of N2 = 767.5 mmHg
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