Question

1.When heated to 350 degrees Celsius at 0.950 atm, ammonium nitrate decomposes to produce nitrogen, water and oxygen gases.

2NH4 NO3 (s)>>>>>> 2N2(g) +4H2O(g) +O2 (g)

How many liters of water vapor are produced when 25.6 grams of NH4NO3 decomposes?

How many grams of NH4NO3 are needed to produce 12.7L of oxygen?

2. A multipaitent hyperbaric chamber has a volume of 3200 L.

At a temperature of 21 degrees Celsius how many grams of Oxygen are needed to give a pressure of 2.7 atm?

3. A balloon is floating around outside your window. The temperature outside is 17 degrees Celsius, and the air pressure is 0.700 atm. Your neighbor who released the balloon, tells you that he filled it with 3.60 moles of gas. What is the volume of gas inside the balloon?

4. A 10.0 L gas cylinder is filled with 9.20 moles of gas. The tank is stored at 1 degree Celsius. What is the pressure in the tank?

5. A 200. L kiln is used for vitrifying ceramics. It is currently operating at 905 degrees Celsius, and the pressure is 1.100 atm. How many moles of air molecules are within the confines of the kiln?

6. A 18.0 L gas cylinder has been filled with 7.60 moles of gas. You measure the pressure to be 3.80 atm. What is the temperature inside the tank?

Answer 1

1. Given equation is:

2NH₄NO₃ (s) 2N₂(g) +4H₂O(g) +O₂ (g)

Mass of NH₄NO₃ = 25.6 g

Molar mass of NH₄NO₃ = 80 g/mol

Moles of NH₄NO₃ = Mass of NH₄NO₃ /Molar mass of NH₄NO₃ =  25.6 g/ (80 g/mol) = 0.32 mol

From reaction, 0.32 mol of NH₄NO₃ produces 2*0.32 = 0.64 mol of H₂O

Pressure(P) = 0.950 atm

Temperature (T) = 350^oC = 623 K

Gas constant (R) = 0.0821L.atm/ (mol.K)

From ideal gas equation,

PV = nRT

V = nRT/V =  0.64 mol*0.0821L.atm/ (mol.K)*623 K / 0.950 atm = 34.46 L

Agian, Volume of O₂ = 12.7 L

Moles of O₂(n )= PV/RT = 0.950 atm *12.7 L/(0.0821L.atm/ (mol.K)*623 K) = 0.236 mol

1 mol of O₂ produced from 2 mol of NH₄NO₃

Thus, MOles of NH₄NO₃ = 0.472 mol

Molar mass of NH₄NO₃ = 80.052 g/mol

Thus, Mass of NH₄NO₃ = Moles* molar mass = 0.472 mol*80.052 g/mol = 37.78 g

2.volume (V) = 3200 L.

Pressure (P) = 2.7 atm

Temperature (T) = 21^oC= 294 K

Gas constant (R) = 0.0821L.atm/ (mol.K)

n =PV/RT = 2.7 atm* 3200 L./(0.0821L.atm/ (mol.K)*294 K) = 357.95 mol

Molar mass of O₂ = 32 g/mol

Mass of O₂ = 357.95 mol *32 g/mol = 11454.42 g

3. T = 17 ^oC = 290 K

P = 0.700 atm

n= 3.60 moles

V = nRT/V = 3.60 moles*0.0821L.atm/ (mol.K)*290 K / 0.700atm = 122.45 L

4. V= 10.0L

n= 9.20 mol

T = 1^oC = 274 K

P = nRT/V = 9.20 mol*0.0821L.atm/ (mol.K)*274 K/ 10.0L = 20.70 atm

5. V = 200 L

t = 905^oC = 1178 K

p = 1.1 atm

n = 1.1 atm*200 L/ ( 1178 K*0.0821L.atm/ (mol.K)) = 2.27 mol

6. V= 18.0 L

n = 7.60 mol

P = 3.80 atm

T = 3.80 atm*18.0 L/(7.60 mol*0.0821L.atm/ (mol.K)) = 109.6 K = -163.4 ^oC