Ideal Gas Law 02 (8.81) Part A How many moles of a gas will occupy a volume of 0
ID: 550595 • Letter: I
Question
Ideal Gas Law 02 (8.81) Part A How many moles of a gas will occupy a volume of 0.4745 L at a temperature of 360.6 K and a pressure of 1594 torr? Ideal Gas Law 03 (8.80) Part A If 27.0 g of ammonia gas has a temperature of 23.4 °C and a pressure of 579 mmHg, what is its volume? Dumas' Method 1.369 g of an unknown compound were vaporized Part A The resulting gas occupied a volume of 233.9 mL at a pressure of 718.4 torr and 103.5°C . What is the molar mass of the compound? 4A1(s) + 3O2 (g) 2Al2O3 (s) Part A What volume of oxygen is needed at 16.4 °C and 1.105 atm to completely react with 52.36 g of aluminum? which generates nitrogen gas to fill the air bags within 0.03 second 2NaN3 (s) 2Na(s) + 3N2 (g) Part A How many liters of N2 are produced at 20.8°C and 1.39 atm if the air bag contains 128 g NaN ?Explanation / Answer
PV = nRT
T = 360.6K
V = 0.4745L
P = 1594/760 = 2.097atm
n = PV/RT
= 2.097*0.4745/0.0821*360.6 = 0.034moles
part-A
PV = nRT
n = W/M
PV = WRT/M
W = 27g
M = 17g/mole
T = 23.4+273 = 296.4K
P = 579/760 = 0.76atm
V = WRT/PM
= 27*0.0821*296.4/0.76*17 = 50.85L >>>>answer
part-A
PV = nRT
n = W/M
PV = WRT/M
M = WRT/PV
W =1.369g
V = 0.2339L
P = 718.4/760 = 0.945atm
T = 103.5+273 = 376.5K
M = WRT/PV
= 1.369*0.0821*376.5/0.945*0.2339 = 191.45g/mole
part-A
4Al + 3O2-----------> 2Al2O3
4 moles of Al react with 3 moles of O2
4*27g of Al react with 3*32g of O2
52.36g of Al react with = 3*32*52.36/4*27 = 46.54moles of O2
P = 1.105atm
T = 16.4+273 = 289.4K
PV = nRT
V = nRT/P
= 46.5*0.0821*289.4/1.105 = 1000L
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