AP Chemistry Chapter 5 Textbook Answers – Zumdahl
5.51
STP: T = 273 K and P = 1.00 atm; At STP, the molar volume of a gas is 22.42 L.
2.00 L O2 x 1 mol O2/22.42 L x 4 mol Al/3 mol O2 x 26.98 g Al/1 mol Al = 3.21 g Al
5.53
2 NaN3(s) → 2Na(s) + 3N2(g)
n(N2) = PV/RT = 1.00 atm x 70.0 L/0.08206 L atm/K mol x 273 K = 3.123 mol N2 needed to fill
airbag.
mass NaN3 reacted = 3.12 mol N2 x 2 mol NaN3/3 mol N2 x 65.02 g NaN3/1 mol NaN3 =
135 g NaN3
5.57
CH3OH + 3/2 O2 → CO2 + 2H2O or 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g)
50.0 mL x 0.850 g/1 mL x 1 mol/32.04 g = 1.33 mol CH3OH(l) available
n(O2) = PV/RT = 2.00 atm x 22.8 L/0.08206 L atm/K mol x 300. K = 1.85 mol O2 available
1.33 mol CH3OH x 3 mol O2/2 mol CH3OH = 2.00 mol O2
2.00 mol O2 are required to react completely with all of the CH3OH available. We only have
1.85 mol O2, so O2 is limiting.
1.85 mol O2 x 4 mol H2O/3 mol O2 = 2.47 mol H2O
5.61
molar mass = dRT/P where d = density of gas in units of G/L
molar mass = 3.164 g/L x 0.08206 L atm/K mol x 273.2 K/1.000 atm = 70.98 g/mol
The gas is diatomic, so the atomic mass = 70.93/2 = 35.47. This is chlorine and the identity of
the gas is Cl2.
5.63
d (UF6) = P x (molar mass)/RT = [(745 torr x 1 atm/760 torr) x 352.0 g/mol]/[0.08206 L atm/K
mol x 333 K] = 12.6 g/L