Question #288037
A 1.77×10-1 g sample of HNO3 and a 5.21 g sample of NO react in a closed 8.91 L container at 470 K, according to the following balanced chemical equation:
2HNO3(g) + NO(g) → 3NO2(g) + H2O(l)
Calculate the PNO2 (in atm) in the container after the reaction has gone to completion.
Expert's answer
Moles of HNO3 = 1.77 × 10-1 / 63 =0.003 mol.
Moles of NO = 5.21 / 30 = 0.174 mol.
HNO3 is limiting reagent .
So, moles of NO2 formed = 3× 0.003 / 2 mol.
Moles of NO2 = 0.0045 mol.
From ideal gas equation
P = nRT / V = 0.0045 × 0.0821 × 470 / 8.91
P = 0.02 atm ( APPROX) Answer .
