How many liters of nitric acid with a mass fraction of HNO3 48% can be obtained from 44.8 liters of ammonia?

Given:
w (HNO3) = 48%
V (NH3) = 44.8 L
normal
Find: V (HNO3) -?
Decision:
NH3 + 2O2 = HNO3 + H2O
1) Find v (NH3):
v = V / Vm
v (NH3) = 44.8 L / 22.4 L / mol = 2 mol.
v (NH3) = v (HNO3) = 2 mol, since NH3: HNO3 = 1: 1
2) Find the mass for HNO3:
m (HNO3) = v (HNO3) * M (HNO3)
m (HNO3) = 2 * 63 = 126g.
M (HNO3) = 1 + 14 + 16 * 3 = 48 + 14 + 1 = 63 g / mol
3) m (pure substance HNO3) = w (HNO3) * m (HNO3) = 0.48 * 126 = 60.48 g.
4) v (pure HNO3) = 48/63 = 0.96 mol
5) V (HNO3) = 0.96 mol * 22.4 l / mol = 21.504 l.
ANSWER: V (HNO3) = 21.504 l.