Question

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?

Answer 1

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.