Question 28 of 31intermediate🔧 ApplyNumerical3 marks

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?

Correct Answer

The maximum weight of nitric oxide (NO) that can be obtained is 15.00 g.

Exercise: EXERCISES | Q: 7.25 | (Page Page 20)
For More Understanding

Explanation

To solve this stoichiometry problem, we first need the balanced chemical equation for the oxidation of ammonia. The reaction is: 4NH3 + 5O2 -> 4NO + 6H2O. By calculating the moles of each reactant and comparing them to the stoichiometric ratio, we identify the limiting reagent. Oxygen is the limiting reagent here, which dictates the maximum amount of product formed.

Solution Steps

  1. Step 1: Write the balanced chemical equation The reaction for the first step of Ostwald's process is:

    4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g)

  2. Step 2: Calculate the moles of reactants Moles of NH3=10.00 g17 g/mol=0.588 mol\text{NH}_3 = \frac{10.00 \text{ g}}{17 \text{ g/mol}} = 0.588 \text{ mol} Moles of O2=20.00 g32 g/mol=0.625 mol\text{O}_2 = \frac{20.00 \text{ g}}{32 \text{ g/mol}} = 0.625 \text{ mol}

  3. Step 3: Identify the limiting reagent From the balanced equation, 4 moles of NH3\text{NH}_3 require 5 moles of O2\text{O}_2. For 0.588 mol of NH3\text{NH}_3, O2\text{O}_2 required = 54×0.588=0.735 mol\frac{5}{4} \times 0.588 = 0.735 \text{ mol}. Since we only have 0.625 mol of O2\text{O}_2, oxygen is the limiting reagent.

  4. Step 4: Calculate the mass of nitric oxide (NO) produced 5 moles of O2\text{O}_2 produce 4 moles of NO\text{NO}. So, 0.625 mol of O2\text{O}_2 will produce 45×0.625=0.5 mol\frac{4}{5} \times 0.625 = 0.5 \text{ mol} of NO\text{NO}. Molar mass of NO=14+16=30 g/mol\text{NO} = 14 + 16 = 30 \text{ g/mol}. Mass of NO=0.5 mol×30 g/mol=15.00 g\text{NO} = 0.5 \text{ mol} \times 30 \text{ g/mol} = 15.00 \text{ g}.

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