how many grams of Potassium chlorate is taken such that 5.6 litres of oxygen is liberated at ntp
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okay....here ur answer goes;-
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
m(KClO3) = n(KClO3)*M(KClO3) = 0.155 mol * 12.5 g/mol = 19 g
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
m(KClO3) = n(KClO3)*M(KClO3) = 0.155 mol * 12.5 g/mol = 19 g
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