How many liters of nitric oxide (II) are formed when 93.15 g of lead is completely dissolved in dilute nitric acid?

Let’s implement the solution:
m = 93.15 g. X l. -?
1.3Pb + 8HNO3 (solution) = 2NO + 3Pb (NO3) 2 + 4H2O – OBP, nitric oxide is released (2);
2. Let’s make the calculations:
M (Pb) = 207 g / mol;
M (NO) = 30 g / mol.
3. Determine the amount of the original substance, if the mass is known:
Y (Pb) = m / M = 93.15 / 207 = 0.45 mol.
4. Proportion:
0.45 mol (Pb) – X mol (NO);
– 3 mol – 2 mol from here, X mol (NO) = 0.45 * 2/3 = 0.3 mol.
5. Find the volume of the product:
V (NO) = 0.3 * 22.4 = 6.72 liters.
Answer: nitric oxide is released in a volume of 6.72 liters.