Question

3 Cl 2 (aq) + 6 KOH (aq) ---> KClO 3 (aq) + 5 KCl (aq) + 5 H 2 O (l) Assuming that the process is balanced, what is the rate of consumption of potassium hydroxide if the by-product of potassium chloride is being produced at a rate of 49.4 M/min? help urgent

Answer 1

3 Cl 2 (aq) + 6 KOH (aq) ---> KClO 3 (aq) + 5 KCl (aq) + 5 H 2 O (l)

The rate of the given reaction is

rate= 1/3(-dCl₂/dt )= 1/6(-dKOH/dt )= (dKClO₃/dt )=1/5(dKCl/dt )= 1/5(dH₂O/dt )

where

-sign indicate the rate of disappearance of reactant

+sign indicate the rate of appearance of product

It is given that

(dKCl/dt )= 49.4 M/min

since

1/6(-dKOH/dt )=1/5(dKCl/dt )

=>(-dKOH/dt )= 6/5(dKCl/dt )

=>(-dKOH/dt )= (6/5)*49.4 M/min

=>(-dKOH/dt )= 59.28 M/min

Therefore

the rate of consumption of potassium hydroxide= 59.28 M/min

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