What volume of NO will be released during the interaction of copper with 4.5 mol of dilute nitrate acid?

The decision is made based on the reaction:
3Cu + 8HNO3 = 2NO + 3Cu (NO3) 2 + 4H2O – redox reaction, gaseous nitrogen oxide is released (2);
Let’s calculate the molar masses of substances:
M (HNO3) = 1 + 14 + 16 * 3 = 63 g / mol;
M (NO) = 14 +16 = 30 g / mol;
Determine the amount of mol of nitrogen oxide:
4.5 mol (HNO3) – X mol (NO);
-8 mol – 2 mol hence, X mol (NO) = 4.5 * 2/8 = 1.125 mol;
Let’s calculate the volume of nitric oxide using Avogadro’s law:
V (NO) = 1.125 * 22.4 = 25.2 liters.
Answer: during the reaction, nitrogen oxide with a volume of 25.2 liters was released.