compute the mass in grams of kclo3 necessary to produce 67.2 L of oxygen at STP acc. to the reaction
2kclo3 (s) -----> 2 kcl (s) + 3o2 (g)
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According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and contains avogadro's number 6.023 × 10²³ of particles. Thus 245 grams of KClO₃ necessary to produce 67.2 liters of oxygen at S. T. P. At STP , 1 mol gas = 22.4L .
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