Answer to Question #263443 in General Chemistry for x2+y²

Question

When nitric acid is synthesized from Ammonia, the first step in the process is 4NH3(g) + 5O2(g) + 4NO (g) + 6H2O (g)

What volume of NO at STP can be formed from 1250L of NH3 at 325°C and 4.25 atm assuming temperature and pressure conditions remains constant

Answer 1

Balanced equation:

4NH3 + 5O2 --------> 4NO + 6H2O

Assuming all gases act as an ideal gas, so PV = nRT

Number of Moles of NH3 = PV/RT = (4.25atm × 1250L) / (0.082057 L atm mol-1K-1 × (325+ 273)K)

= 108.26 moles

Assuming O2 is in excess and NH3 is the limiting reagent, number of moles of NO = 108.26 moles

At STP, T=273K, P= 1atm,

V = nRT / P = (108.26×0.082057 × 273) / (1) = 2425.19 L

Hence, volume of NO formed at STP = 2425.19 L