# 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?

May 7, 2017

2.75 moles

#### Explanation:

Here, the equation of the reaction is

$2 K C l {O}_{3} \left(s\right)$ -> $2 K C l \left(s\right) + 3 {O}_{2} \left(g\right)$
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. ($0.85 \cdot x = 3.50$ )

Now, 3 molecules of ${O}_{2}$ are given by 2 molecules of $K C l {O}_{3}$
No. of moles of 1 molecule of ${O}_{2}$ = $\frac{4.12}{3}$
By ratio,
No. of moles of 2 molecules of $K C L {O}_{3}$ that reacted = $4.12 \cdot \frac{2}{3}$ = 2.75 moles

Therefore, 2.75 moles of $K C L {O}_{3}$ is required.