What volume of gas will be released when an excess of copper interacts with 100 ml of a 9.54% nitric acid solution (ρ = 1.057 g / ml)

1) Let’s compose the reaction equation:
3Cu + 8HNO3 = 3Cu (NO3) 2 + 2NO ↑ + 4H2O
2) Calculate the mass and amount of the nitric acid substance:
ρ = m / V
m (solution) = ρ (HNO3) ∙ V (HNO3) = 1.057 ∙ 100 = 105.7 g.
m (HNO3) = (m (solution)) ∙ ω (HNO3) / 100% = 105.7 ∙ 9.54 / 100 = 10.08 g.
n (HNO3) = m (HNO3) / M (HNO3) = 10.8 / 63 = 0.16 mol.
3) Determine the amount of substance and the volume of released gas:
Let’s make the proportion:
n (NO) / 2 = n (HNO3) / 8
n (NO) = 0.16 ∙ 2/8 = 0.04 mol
V (NO) = n (HNO3) ∙ Vm = 0.04 • 22.4 = 0.896 l.
Answer: 0.896 l