Calculate the mass of carbon monoxide (IV) that will be released during firing 152 g of limestone containing 9% impurities.

Given:
m s (CaCO3) = 152 g
w p (CaCO3) = 9%
Find: m (CO2)
Decision:
CaCO3 = CaO + CO2
w o.c. (CaCO3) = 100% -9% = 91% = 0.91
m in (CaCO3) = w h.v * m s = 0.91 * 152 g = 138.32 g
n (CaCO3) = m / M = 138.32 g / 100 = 1.4 mol
n (CaCO3): n (CO2) = 1: 1
n (CO2) = 1.4 mol
m (CO2) = n * M = 1.4 mol * 44 g / mol = 61.6 g
Answer: 61.6 g