Let’s write down the reaction scheme and arrange the coefficients:
CaCO3 + 2HCl = CaCl2 + CO2 + H2O
Determine the mass of pure calcium carbonate m (CaCO3) = m (v-va) * (1-w), where m (v-va) is the mass of the contaminated substance, w is the mass fraction of impurities
m (CaCO3) = 200 * (1 – 0.15) = 170 g
Next, we calculate the amount of substance n (CaCO3) = m (CaCO3) / M (CaCO3), where M (CaCO3) = 100 g / mol is the molar mass of calcium carbonate
n (CaCO3) = 170/100 = 1.7 mol
It can be seen from the reaction equation that the amount of carbon dioxide substance n (CO2) = n (CaCO3) = 1.7 mol.
Let us determine the volume of carbon dioxide V (CO2) = Vm * n (CO2), where Vm = 22.4 l / mol is the molar volume.
V (CO2) = 22.4 * 1.7 = 38.08 l
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