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

If N2O5 is decomposing with an instantaneous rate of 7.81 mol/L·s, what is the instantaneous rate of formation of NO2?
I don't know where to begin

Answer :

taskmasters
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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