What mass of potassium permanganate is required to obtain 10 mol of oxygen?
| November 27, 2020 | education
2KMnO4 = K2MnO4 + MnO2 + O2
m = n * M (Mr)
Mr (KMnO4) = 39 + 55 + 4 * 16 = 158
M (KMnO4) = 158 grams / mol
n (KMnO4) = 10 * 2 = 20 mol, since according to the reaction condition it is 2 times more
m (KMnO4) = 20 * 158 = 3160 grams = 3.160 kilograms
Answer: 3.16 kilograms.
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