0.05 gm of commercial sample of KCIO:on decomposition liberated just sufficient oxygen for complete
oxidation of 20 ml CO at 27°C and 750 mm pressure. Calculate % KCIO3 in sample
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Answer:
Explanation:
The number of moles of CO at 27∘C and 75 mm Hg pressure
PV=nRT
n=PVRT=750760×201000×10.082×300=8.023×10−4mol
∴ number of moles O2=8.023×10−42
=4.0115×10−4 mol
2KCIO3→2KCl+3O2
2CO+O2→2CO2
Weight of KCIO3 (pure) =4.0115×10−4×2×122.53
=0.03276g
% of purity=0.032760.05×100=65.52
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