Ammonia and Oxygen

Question

A mixture of ammonia and oxygen is prepared by combining 300 mL of NH3 (measured at 759 torr and 28 Celsius) with 220 mL of O2 (measured at 780 torr and 50 Celsius). How many milliliters of N2 (measured at 740 torr an 100 Celsius) could be formed if the following reaction occurs?

4NH3 (g) + 3O2 (g) → 2N2 (g) + 6H2O (g)

Answer 1

Use ideal gas law equation or PV = nRT, with P being pressure in atm, V being volume in liters, n being moles of gas, R being a constant of .08206, and T being temperature in Kelvin.

Use ideal gas law equation and given volume (convert from mL to L), pressure (convert torr to atm using torr/760 = atm), and temperature (convert Celcius to Kelvin using C + 273 = K) to find moles of NH3:

(.99868 atm)(.3L) = n (.08206)(301)

= .0121297 moles NH3

Use ideal gas law equation and given volume, pressure, and temperature using previous conversions to find moles of O2:

(1.0263 atm)(.22L) = n (.08206)(323)

= .00852 moles O2

Now use moles of each molecule to find moles of N2 to determine limiting reactant:

.0121297 moles NH3 x 2 moles N2 / 4 moles NH3

= .008086 moles N2 produced from NH3

.00852 moles O2 x 2 moles N2 / 3 moles O2

=.005679 moles N2 produced from O2 . <--- O2 is limiting as it can produce the least amount of product.

Now use ideal gas law, with theoretical moles of N2 produced (moles of N2 produced from limiting reactant) to find volume in mL of N2 produced. Use provided temperature and pressure values (be sure to convert).

(.97368 atm)(V) = (.005679 moles N2)(.08206)(373 K)

= .1785 L or 178.5 mL N2