A sample of KCIO3 on decomposition yielded 448 mL of oxygen gas at S
TP.
Calculate (1) Weight of oxygen product, (ii) Weight of KCIO3 originally taken, and (ii) Weight of KCI
produced. (K = 39, CI = 35.5 and 0 = 16)
Answers
Answer:
2KClO
3
Δ
3O
2
+2KCl
2mole3mole2mole
Number of moles of O
2
at S.T.P.=
22400
V(ml)
=
22400
448
=0.02mole
Mass of 1 mole of O
2
= 32 grams
Mass of 0.02 mole of O
2
=32× 0.02 = 0.64 grams
Mass of KClO
3
=moles(n)×molecular weight=2× 122.5 = 245 grams
Weight of KClO
3
to produce 3 moles of O
2
= 245 grams
Weight of KClO
3
to produce 0.02 moles of O
2
=
3
245×0.02
= 1.63 grams
Mass of KCl produced = Mass of KClO
3
-Mass of O
2
produced =1.63−0.64=0.993 grams
Weight of O
2
produced = 0.64 grams
Weight of KClO
3
originally taken = 1.63 grams
Weight of KCl produced = 0.993 grams
Answer:
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