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If the rate of appaerance of O2 in the reaction: 2O3 (g)-->3O2(g) is .250 M/s over...

If the rate of appaerance of O2 in the reaction: 2O3 (g)-->3O2(g) is .250 M/s over the first 5.50 , how much oxygen will form during this time. Please explain!

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Answer #1

rate = 0.250 M / s

time = 5.50 sec

cocentration of O2 produced = rate x time

                                              = 0.250 x 5.50

                                             = 1.375 M

O2 produced = 1.38 M

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