Find the mass of salt and the volume of gas formed by the interaction of chromium with concentrated

Find the mass of salt and the volume of gas formed by the interaction of chromium with concentrated nitric acid, if it is known that 7.2 g of water was formed.

Given:
m (H2O) = 7.2 g
To find:
m (Cr (NO3) 3)
V (NO2)
Decision:
Cr + 6HNO3 = Cr (NO3) 3 + 3NO2 + 3H2O
n (H2O) = m / M = 7.2 g / 18 g / mol = 0.4 mol
n (H2O): n (Cr (NO3) 3) = 1: 1
n (Cr (NO3) 3) = 0.4 mol
n (H2O): n (NO2) = 1: 3
n (NO2) = 0.4 mol * 3 = 1.2 mol
m (Cr (NO3) 3) = n * M = 0.4 mol * 238 g / mol = 95.2 g
V (NO2) = n * Vm = 1.2 mol * 22.4 L / mol = 26.88 L
Answer: 95.2 g; 26.88 l