2.50g of KClO3 decomposes by the equation: 2KClO3=2KCl+3O2. What volume of O2 is
ID: 720458 • Letter: 2
Question
2.50g of KClO3 decomposes by the equation: 2KClO3=2KCl+3O2. What volume of O2 is produced at 25 degree C and a pressure of 750 mmHg?Explanation / Answer
2KClO3=======> 2KCl + 3O2 2.50g of KclO3 = 2500 g KclO3 Moles = mass/molar mass Moles of KclO3 = 2500 g /122.55 g/mol=20.40 mol KclO3 Here the molar ratio is 2 KclO3 : 3 O2 20.40 mol KclO3 * ( 3 mol O2 /2 mol KclO3) 30.60 mol O2 Now pressure (P) = 750 mm Hg = 0.986 atm : temperature (T) = 25 C =298 K : R= gas constant = 0.0821 L atm /mol K From ideal gas law PV=nRT V =nRT/P = (30.60 * 0.0821*298)/0.989 = 756.98 L O2
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