How many grams of KClO3 is taken such that 5.6 litres of oxygen is liberated at NTP?
NTP - Normal Temperature and Pressure - is defined as air at 20oC (293 K) and 1 atm ( 101.325 kPa). The KClO3 decomposition: 2 KClO3 = 2 KCl + 3 O2 The amount of oxygen is pV = nRT n(O2) = pV / RT = 1 atm * 5.6 L / (0.08206 L*atm/K*mol * 293 K) = 0.233 mol n(KClO3) = (2/3)*n(O2) = 0.233*(2/3) = 0.155 mol The mola mass of KCLO3 is M(KClO3) = 122.5 /mol The mass of KClO3 is
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