The Composition of Potassium Chlorate Part A: Mass Percent of Oxygen in Potassiu

Question

The Composition of Potassium Chlorate Part A: Mass Percent of Oxygen in Potassium Chlorate Experimental Data Sample 1 2418 (a) Mass of crucible + lid (b) Mass of crucible, lid+ KCIO (c) Mass of crucible, lid+ residue after 1" heating (d) Mass of crucible, lid + residue after 2d heating 13.048 13, o 69. Data Analysis . Use your data to determine the experimental mass percent of oxygen in KCIO, Show your work clearly for each step in the table below. Sample 2 Mass oforiginal KCIO, sample | 13 H 47-1 2.4 8- 0.989 a CIO 13.04-12.478 Mass of KCI residue 0.6 68 KCi nas 0.989-0.5C8 Mass of Oxygen released Mass Percent of Oxygen in KCio, | 0,421 0.989 x100-42.57% Page I of 3

Explanation / Answer

1) The experimental mass percent of O2 is higher than theoretical yield.

Mass of O2=Mass of KClO3(original)-massof KCl(residue) ,If residue is low in mass then mass of O2 is higher,and this happens when some of KCl residue gets spattered out of the crucible.

(high mass of O2/mass of KClO3)*100=high mass % of O2

2)potassium chlorite,KClO2 -----heat-->KCl +O2

potassium chlorite will also give out O2 gas

O2 produced in for 1mol KClO3(122.55g/mol)=3/2*mol O2=3/2*32g/mol=48 g

2KClO3 --heat--->2KCl(s) +3O2 (

O2 produced in for 1mol KClO2(106.546g/mol)=1 mol O2=32 g

O2 produced for 1 g KClO2=32/106.546=0.3 g

So,1 g KClO3 produces =48/122.55=0.392 g O2 (higher amount than KClO3)

thus decomposition of mixture of KClO3+KClO2 will givea lower yield

2)

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