Consider the following reaction: 4nh3(g) 7o2(g) → 4no2(g) 6h2o(l) consider an experiment in which you react ammonia and oxygen. at the end of the experiment, you find that you produced 27.0 g of water, and 8.52 g of ammonia is left over. calculate the initial mass of ammonia. assume the reaction went to completion.

Initial mass of ammonia: 25,537g NH3

Explanation:

4NH3(g) 7O2(g) → 4NO2(g) 6H2O(l)

We have 4 Nitrogen atoms on both sides

12 Hydrogen atoms on both sides

14 Oxygen atoms on both sides

Final Mass of H2O = 27g  and Molecular Weigth = 18,015g/mol

Left Mass of NH3 = 8,52g and Molecular Weight= 17,031 g/mol

27g H2O  x  1mol H20/ 18,015 gH20  x 4mol NH3/ 6 mol H20 x 17,031g NH3 / 1mol NH3 = 17,017g NH3

17,017g NH3 were consumpted to generate 27 g of H2, however, lets remember that there is NH3 left, that must be added to finally discover the original amount

17,017g + 8,52g = 25,537g NH3

what's the

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :