What volume of chlorine is released when 63.2 g of potassium permanganate interacts with an excess
| December 8, 2020 | education
What volume of chlorine is released when 63.2 g of potassium permanganate interacts with an excess of concentrated hydrochloric acid, if the yield of the reaction product is 87.5%?
2KMnO4 + 16HCl = 2MnCl2 + 5Cl2 ↑ + 2KCl + 8H2O
n (KMnO4) = 63.2 / 158 = 0.4 mol
n (Cl2) = (0.4 * 5/2) * 0.875 = 0.875 mol
V (Cl2) = 0.875 * 22.4 = 19.6 L
Answer: 19.6 L
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