If you place 10.0 L of propanol (C3H8O) in a sealed room that is 3.00 m long, 2.

Question

If you place 10.0 L of propanol (C3H8O) in a sealed room that is 3.00 m long, 2.75 m wide, and 2.50 m high. The vapor pressure of propanol is 10.7 torr at 25 °C, and the density of the liquid at this temperature is 0.804 g/mL. Treat the room dimensions as exact numbers. If some liquid remains, how much will there be? ____ L Given: 1m3 = 1000L ; 1 atm = 760 torr ; K = 273.15 + C ; R = 0.082057 L atm/(mol K) = 8.314 J / (mol K)

Methylamine burns in oxygen according to the following equation: 4 CH5N + 9 O2 ® 4 CO2 + 10 H2O + 2 N2 Report your answers to parts (a) and (b) to 3 significant figures. (a) How many liters of O2 at 57. °C and 0.610 atm will be needed to burn 8.98 L of Methylamine at 57. °C and 0.610 atm? _____L O2

Explanation / Answer

PV = nRT

P = 10.7/760 atm = 0.0141 atm (Pressure exerted by propanol gas)

V = 3*2.75*2.50 m^3 = 20.625 m^3 =20.625*1000 L =20625 L

T = 25 oC = 298.15 K

R = 0.082 L-atm/m/K

n=?

So, n = PV/RT = 0.0141*20625/(0.082*298.15) mol = 11.895 mol

10 L of propanol = 1000*10*0.804 g (density = 0.804 g/mL) = 8040 g = 8040/60 mol (molar mass of propanol = 60 g/mol) = 134 mol

So, out of 134 mol of propanol, 11.895 mol vaporize, therefore remaining (134-11.895) mol will be in liquid form = 122.105 mol = 7326.3 g =9112.3 mL = 9.11 L

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