Question

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. \(4NH_{3}(g)+5O_{2} --> 4NO+6H_{2}O(g)\) What is the maximum mass of H2O that can be produced by combining 72.0 g of each reactant?

Answer 1

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

68 g              160 g                      120 g            108 g

72.0 g            72.0 g                                          ??

here

68 g NH3 required ----------------> 160 g O2

72 g NH3 required ------------->          ?? O2

mass of O2 = 72 x 160 / 68 = 169.4 g

but here we have only 72.0 g . so

limiting reagent is O2. so product formed according to that.

160 g O2 gives ----------------> 108 g H2O

72.0 g O2 -------------------->   ?? H2O

mass of H2O produced = 72 x 108 / 160 = 48.6 g

maximum mass of H2O that can be produced = 48.6 g