A sample of gas contains 0.1200 mol of NH3(g) and 0.1500 mol of O2(g) and occupies a volume of 13.0 L. The following reaction takes place: 4NH3(g) + 5O2(g)4NO(g) + 6H2O(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.

14.44 Liters

Explanation:

4NH3(g) + 5O2(g) > 4NO(g) + 6H2O(g)

From the equation of the reaction, the mole ratio of NH3 to NO2 = 1:1

Therefore 0.12 moles of NH# produces 0.12 moles of NO

Also from the equation of reaction, mole ratio of NH3 to H20 is 4:6

Therefore, 0.12 moles of NH3 will produce  0.12 * 6/ 4 moles of H2O =0 0.18 moles of H2O

Total moles of gases produced = 0.30 mole

From Avogadro's law:

v1/n1 = v2/n2

v1 = 13.0 L

n1 = 0.12 + 0.15= 0.27 moles

v2 = ? =

n2 = 0.30

V2 = v1n2/n1

v2 = 13.0 * 0.30/0.27

V2 = 14.44 Liters

i think that the answer is c