Determine the volume of chlorine that will be released during the interaction of 36.5% hydrochloric acid, weighing 200 g.

univerkov | December 8, 2020 | education

4HCL + O2 = 2Cl2 + 2H2O
m HCl = 200 * 0.365 = 73 g
n HCl = 73 / 36.5 = 2 mol
n Cl2 = 2 mol
V Cl2 = 22.4 * 1 = 22.4 l
We get the answer: 73 g, 2 mol, 2 mol, 22.4 L.

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