Determine the volume of carbon dioxide that will be released during the interaction of 120 g of chalk with hydrochloric acid, if the practical reaction yield is 70%?
CaCO3 + 2HCL = CaCl2 + H2O + CO2
n (caco3) = m / M = 120/100 = 1.2 mol
The level of the reaction shows that n caco3: n co2 as 1: 1, which means n co2 = 1.2 mol
V (CO2) = n * Vm; V (CO2) = 1.2 * 22.4 = 26.88L (this is theoretical V)
Looking for a practical V
V = 22.88L * 0.7 = 18.816L
Answer: V (CO2) = 18.816L
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