What volume of CO2 will be released when 45 g of limestone containing 8% impurities interacts with nitric acid?

Given:
m s (CaCO3) = 45 g
w p (CaCO3) = 8%
To find:
V (CO2)
Decision:
CaCO3 + 2HNO3 = Ca (NO3) 2 + H2O + CO2
w p.w (CaCO3) = 100% -8% = 92% = 0.92
m in (CaCO3) = w h.v * m s = 0.92 * 45 g = 41.4 g
n (CaCO3) = m / M = 41.4 g / 100 g / mol = 0.414 mol
n (CaCO3): n (CO2) = 1: 1
n (CO2) = 0.414 mol
V (CO2) = n * Vm = 0.414 mol * 22.4 l / mol = 9.27 l
Answer: 9.27 L