Question

The given reaction is an important step in Ostwald's method for manufacturing of HNO3. If we start with 6.8 g of NH3 and 40 g of O2, then what mass % of excess reagent will be left behind ?

Answer 1

The given reaction in the Ostwald process for the manufacture of nitric acid is:

4NH₃+5O₂→4NO+6H₂O

Step 1: Calculate the moles of reactants

• Molar mass of NH₃ = 17 g/mol
  Moles of NH₃ = 6.817=0.4\(\frac{6.8}{17}\) = 0.4 moles

• Molar mass of O₂ = 32 g/mol
  Moles of O₂ = 4032=1.25\(\frac{40}{32}\) = 1.25 moles

Step 2: Determine the limiting reagent

From the balanced reaction:

• 4 moles of NH₃ react with 5 moles of O₂
• 1 mole of NH₃ requires 54=1.25\(\frac{5}{4} = 1.25\) moles of O₂

For 0.4 moles of NH₃, required O₂ = 0.4×1.25=0.50.4\( \times\) 1.25 = 0.5 moles

Since we have 1.25 moles of O₂ available, and only 0.5 moles are needed, NH₃ is the limiting reagent, and O₂ is in excess.

Step 3: Calculate the remaining excess O₂

Used O₂ = 0.5 moles
Initial O₂ = 1.25 moles
Remaining O₂ = 1.25 - 0.5 = 0.75 moles

Mass of remaining O₂ = 0.75×32=240.75\( \times\) 32 = 24 g

Step 4: Calculate mass % of excess O₂ left

\(\text{Mass % of excess reagent left} = \left( \frac{\text{Mass of excess reagent left}}{\text{Initial mass of excess reagent}} \right) \times 100 =(2440)×100=60%= \left( \frac{24}{40} \right) \times 100 = 60\%\)

Final Answer:

Mass % of excess reagent (O₂) left = 60%.