For the reaction: N2O5(g) \longrightarrow&longrightarrow; 2NO2(g) + 1/2O2(g)

1 answer:

5 0

The given equation from the problem above is already balance,
N2O5 ---> 2NO2 + 0.5O2
Since, in every mole of N2O5 consumed, 2 moles of NO2 are formed, we can answer the problem by multiplying the given rate, 7.81 mol/L.s with the ratio.
(7.81 mol/L.s) x (2 moles NO2 formed/ 1 mole of N2O5 consumed)
= 15.62 mol/L.s
The answer is the rate of formation of NO2 is approximately 15.62 mol/L.s.

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