When 22.8 g of copper interacted with nitric acid, 6.86 liters of nitric oxide (4)   were released. Calculate the volume

When 22.8 g of copper interacted with nitric acid, 6.86 liters of nitric oxide (4)   were released. Calculate the volume fraction of the yield of nitrogen oxide (4) from the theoretically possible.

Given:
m (Cu) = 22.8g
V (NO2) = 6.86 l
Decision:
Cu + 4HNO3 = Cu (NO3) 2 + 2NO2 + 2H2O
Let’s calculate the amount of copper substance:
n (Cu) = m (Cu) / M (Cu) = 22.8g / 64g / mol = 0.3 mol
According to the reaction equation:
n (NO2) = 2n (Cu) = 0.6 mol
Let’s calculate the theoretical volume of nitric oxide:
V (NO2) = Vm * n (NO2) = 22.4 l / mol * 0.6 mol = 13.44 liters
Let’s calculate the product yield:
η = V (practical) / V (theoretical) = 6.86 l / 13.44 l = 0.5 * 100% = 50%