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. What is the maximum mass of H2O that can be produced by combining 54.1 g of each reactant?
4N3(g)+5O2(g) --> 4NO(g)=6H2O (g)
4 NH3 (g) + 5 O2 (g) = 4 NO (g) + 6 H2O (g)
Mass of NH3 = 54.1 g.
Molar mass of NH3 = 17.0 g/mol
Moles of NH3 = mass / molar mass = 54.1 / 17.0 = 3.18 mol
Mass of O2 = 54.1 g.
Molar mass of O2 = 32.0 g/mol
Moles of O2 = 54.1 / 32.0 = 1.69 mol
from the balanced equation,
4 mol of NH3 needs 5 mol O2
then,
3.18 mol of NH3 needs 5 * 3.18 / 4 = 3.98 mol of O2
But it is given only 1.69 mol of O2.
Hence, O2 is limiting reagent.
From the balanced equation,
5 mol of O2 forms 6 mol of H2O
Then, 1.69 mol of O2 forms 6 * 1.69 = 10.1 mol of H2O
Therefore,
Mass of H2O produced = moles * molar mass = 10.1 * 18.0 = 182. g.
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