Determine the volume of nitric oxide (2) formed by the interaction of 128 grams of copper with dilute HNO3.

Let’s execute the solution:
1. We write down the equation according to the problem statement:
m = 128 g. X l. -?
3Cu + HNO3 (solution) = 2NO + 3Cu (NO3) 2 + 4H2O – OBP, nitrogen oxide is released (2);
2. We make calculations:
M (Cu) = 63.5 g / mol;
M (NO) = 30 g / mol.
3. Find the number of moles of the starting material:
Y (Cu) = m / M = 128 / 63.5 = 2 mol.
4. The proportion corresponds to:
2 mol (Cu) –X mol (NO);
3 mol            – 2 mol from here, X mol (NO) = 2 * 2/3 = 1.3 mol.
5. Determine the volume of the product:
V (NO) = 1.3 * 22.4 = 29.12 liters.
Answer: Nitric oxide was obtained with a volume of 29.12 liters.