# How many moles of potassium chlorate must be used to produce 6 moles of oxygen gas?

Dec 21, 2016

Following the equation:
$2 K C l {O}_{3} \implies 2 K C l + 3 {O}_{2}$

where, $K C l$ = Potassium Chloride,
${O}_{2}$ = Oxygen Gas
$K C l {O}_{3}$ = Potassium Chlorate

So, 6 moles of Oxygen Gas ${O}_{2}$ has been given to you.
To calculate the number of moles for potassium chlorate $K C l {O}_{3}$, use the mole ratio.

Looking closely at the equation,
If 3 moles of ${O}_{2}$ gives you 2 moles of $K C l {O}_{3}$ because $3 \div 3 \times 2$,
then 6 moles of ${O}_{2}$ must give you:
$6 \div 3 \times 2$ = 4 moles of $K C l {O}_{3}$

To find the number of moles for another compound in the same equation:
number of moles given to you from compound 1 $\div$ the front number for compound 1 $\times$ the front number for another compound.

Therefore, 4 moles of potassium chlorate must be used to produce 6 moles of oxygen gas.