Question #12522

How many grams of KClO3 is taken such that 5.6 litres of oxygen is liberated at NTP?

Expert's answer

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

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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