Question

In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen ?

Answer 1

The maximum weight of nitric oxide produced in Ostwald's process from 10.00 g of ammonia and 20.00 g of oxygen is 15.005 grams, with oxygen being the limiting reactant.

Step-by-step explanation:

In Ostwald's process for the manufacture of nitric acid, we can calculate the maximum weight of nitric oxide (NO) that can be obtained from the reaction: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)

First, we need to determine the limiting reactant, which is the reactant that will be completely consumed and limit the amount of product formed.

To find the limiting reactant, we calculate the molar mass of NH3 which is 17.031 g/mol, and of O2 which is 32.00 g/mol.

Thus, starting with 10.00 g of NH3, we have 10.00 g / 17.031 g/mol = 0.587 mol of NH3.

For the 20.00 g of O2, we have 20.00 g / 32.00 g/mol = 0.625 mol of O2.

According to the balanced equation, the molar ratio between NH3 and O2 is 4:5.

Therefore, theoretically, 0.587 mol of NH3 would require 0.587 mol * (5/4) = 0.734 mol of O2.

Since we only have 0.625 mol of O2, O2 is the limiting reactant.

Now, we've established that oxygen is the limiting reactant.

Using the stoichiometry of the reaction and the amount of limiting reactant, we calculate the moles of NO produced. Since 5 mol of O2 produce 4 mol of NO, then 0.625 mol of O2 will produce 0.625 mol * (4/5) = 0.500 mol of NO.

The molar mass of NO is 30.01 g/mol.

Thus the maximum weight of nitric oxide produced is 0.500 mol * 30.01 g/mol = 15.005 g of NO.