Question

Nitric acid is produced commercially by the Ostwald process, represented by the following equations. 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) 2 NO(g) + O2(g) 2 NO2(g) 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) What mass in kg of NH3 must be used to produce 7.6 ✕ 106 kg HNO3 by the Ostwald process, assuming 100% yield in each reaction?

Answer 1

To produce 7.6 × 106 kg of HNO₃ via the Ostwald process with 100% yield, approximately 2.04 × 106 kg of NH₃ is required.

Step-by-step explanation:

The student asked about the mass of NH₃ required to produce 7.6 × 106 kg of HNO₃ using the Ostwald process, assuming a 100% yield. We use stoichiometry to find the relationship between NH₃ and HNO₃ from the balanced chemical equation provided:

1. 4NH₃(g) + 5O₂(g) ⇒ 4NO(g) + 6H₂O(g)
2. 2NO(g) + O₂(g) ⇒ 2NO₂(g)
3. 3NO₂(g) + H₂O(l) ⇒ 2HNO₃(aq) + NO(g)

From the third equation, 3 moles of NO₂ produce 2 moles of HNO₃. The molar mass of NH₃ is 17.03 g/mol and the molar mass of HNO₃ is 63.01 g/mol. The required mass of NH₃ can be calculated as follows:

(7.6 × 106 kg HNO₃) × (1,000,000 g/kg) × (1 mol HNO₃/63.01 g HNO₃) × (1 mol NO₂2/1 mol HNO₃) × (4 mol NH₃/4 mol NO₂) × (17.03 g NH₃/1 mol NH₃) × (1 kg/1,000,000 g) = Mass of NH₃ in kg

After completing the calculations, the mass of NH₃ needed is approximately 2.04 × 106 kg.

Answer 2

The answer to your question is: 1538095.2 kg of NH3

Step-by-step explanation:

MW HNO3 = 63 kg

MW NO2 = 46 kg

3 NO2(g) + H2O(l)--- 2 HNO3(aq) + NO(g)

3(46) kg-------------- 2(63) kg

x --------------- 7600000 kg

x = 7600000 x 138/126 = 8323809.5 kg og NO2

MW NO = 30

2 NO(g) + O2(g)---2 NO2(g)

2(30) ------------------2(46)

x ---------------- 8323809.5 kg

x = 8323809.5 x 60/92 = 5428571.4 kg of NO

MW NH3 = 17 kg

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

4(17) -------------------- 4(30)

x ----------------------- 5428571.4

x = 5428571.4 x 34 / 120

x = 1538095.2 kg of NH3