What volume of carbon monoxide (4) is formed by the interaction of 60 g of marble, containing 8 percent

What volume of carbon monoxide (4) is formed by the interaction of 60 g of marble, containing 8 percent of non-carbonate impurities, with an excess of nitric acid. A detailed solution to the problem is possible, the answer should be 12.4 liters.

Given:
m mixture (CaCO3) = 60 g
w approx (CaCO3) = 8%
To find:
V (CO2)
Decision:
CaCO3 + 2HNO3 = Ca (NO3) 2 + H2O + CO2
w is clean. islands (CaCO3) = 100% -8% = 92% = 0.92
m clean in islands (CaCO3) = w * m mixture = 0.92 * 60 g = 55.2 g
n (CaCO3) = m / M = 55.2 g / 100 g / mol = 0.552 mol
n (CaCO3): n (CO2) = 1: 1
n (CO2) = 0.552 mol
V (CO2) = n * Vm = 0.552 mol * 22.4 l / mol = 12.4 l
Answer: 12.4 L