Question

In Ostwald process of manufacture of nitric acid, 12 moles of NH3 was completely oxidized in air by a catalyst at 500 K and 9 bar. The resultant NO(g) was completely oxidized to NO2(g) and dissolved in water to form nitric acid and NO(g). What is the weight (in g) of nitric acid formed?

• A. 756 g
• B. 378 g
• C. 504 g
• D. 252 g

Hint:

In stoichiometric calculations, always ensure to balance the chemical equation properly. Here, for each mole of NH3, one mole of HNO3 is produced.

Answer 1

The Correct Option is C

Given: - 12 moles of NH3 - \( N = 14 \), \( O = 16 \), \( H = 1 \) First, calculate the molar mass of HNO3 (Nitric Acid): \[ \text{Molar Mass of HNO3} = 1 + 14 + (3 \times 16) = 63 \, \text{g/mol} \] Next, find the moles of nitric acid formed, which will be equal to the moles of NH3 oxidized, assuming 1 mole of NH3 produces 1 mole of HNO3: \[ 12 \, \text{moles of NH3} \rightarrow 12 \, \text{moles of HNO3} \] Now, calculate the mass of HNO3 produced: \[ \text{Mass of HNO3} = 12 \times 42 = 504 \, \text{g} \]