The decomposition of KClO3(s) into KCl(s) and O2(g) with an 85% yield. How many moles of KClO3 must be used to generate 3.50 moles of O2?

1 Answer
May 7, 2017

Answer:

2.75 moles

Explanation:

Here, the equation of the reaction is

#2KClO_3(s) # -> #2KCl (s) + 3O_2(g) #
Now as you see, the yield of the reaction is 85% which means that out of the number of moles of the products to be given out, only 80% is obtained.
In order to obtain 3.50 moles of #O_2#, which is 85% of the no. of moles which ought to be obtained.
So, if the reaction gave 100% of the yield, no. of moles of #O_2# would be 4.12 moles. (#0.85*x = 3.50 # )

Now, 3 molecules of #O_2# are given by 2 molecules of #KClO_3#
No. of moles of 1 molecule of #O_2# = #4.12/3#
By ratio,
No. of moles of 2 molecules of #KCLO_3 # that reacted = #4.12*2/3# = 2.75 moles

Therefore, 2.75 moles of #KCLO_3 # is required.