How many liters of nitrogen (IV) will be released when 63 grams of 60% nitric acid solution interacts with copper?

Given:
ω (HNO3) = 60%
m solution (HNO3) = 63 g

To find:
V (NO2) -?

Decision:
1) Cu + 4HNO3 => Cu (NO3) 2 + 2NO2 ↑ + 2H2O;
2) m (HNO3) = ω (HNO3) * m solution (HNO3) / 100% = 60% * 63/100% = 37.8 g;
3) M (HNO3) = Mr (HNO3) = Ar (H) * N (H) + Ar (N) * N (N) + Ar (O) * N (O) = 1 * 1 + 14 * 1 + 16 * 3 = 63 g / mol;
4) n (HNO3) = m (HNO3) / M (HNO3) = 37.8 / 63 = 0.6 mol;
5) n (NO2) = n (HNO3) * 2/4 = 0.6 * 2/4 = 0.3 mol;
6) V (NO2) = n (NO2) * Vm = 0.3 * 22.4 = 6.72 liters.

Answer: The volume of NO2 is 6.72 liters.