calculate the mass of potassium chlorate required to liberate 6.72 dm3 of oxygen at STP , molar mass of KCl3 is 112,5 g/mol.

Asked by  | 24th May, 2012, 03:45: PM

Expert Answer:

1 dm3 = 1 L
 2KClO3  ? 2KCl + 3O2
From equation ,
22.4 x 3 dm3  liberate  oxygen by  = 2 x 112.5 g KClO3
1   liberate  oxygen by           = 112.5 x2 / 3 x 22.4
6.72 dm3 ' liberate  oxygen by      '= 6.72  x 112.5 x 2 / 3 x 22.4
                                                  = 225 g

Answered by  | 31st May, 2012, 10:58: AM

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