# How many moles of oxygen gas will be formed from 6.45 g of potassium chlorate?

#### Answer:

number of moles in $K C l {O}_{3}$ $\div 2 \times 3$ = ??

#### Explanation:

This is your equation:
$2 K C l + 3 {O}_{2} \implies 2 K C l {O}_{3}$

Assume that:
n = number of moles
m = mass of substance
M = molar mass

$n = m \div M$

The mass (m) of $K C l {O}_{3}$ has been provided for you: 6.45 grams.

Now you need to find the molar mass (M) for $K C l {O}_{3}$. You know that $K C l {O}_{3}$ is composed of K, Cl and O. If you refer to your periodic table, find the molar mass of each atom and add them up.
K = 1 $\times$ 39.1 = 39.1 g/mol
Cl = 1 $\times$ 35.5 = 35.5 g/mol
O = 3 $\times$ 16.0 = 48.0 g/mol
Now add them all up.
39.1 + 35.5 + 48.0 = 122.6 g/mol is the molar mass (M) of $K C l {O}_{3}$.

Now you can find the number of moles (n) in $K C l {O}_{3}$.
$n = 6.45 \div 122.6$ = 0.05261011419 moles
It's best not to round yet just so it won't give you inaccurate answer at the end.

If you look closely in your equation, you can see that the mole ratio between $K C l {O}_{3} : {O}_{2}$ is $2 : 3$.

If 2 moles of $K C l {O}_{3}$ gives you 3 moles of ${O}_{2}$,
then 0.05261011419 moles of $K C l {O}_{3}$ will gives you:

[$0.05261011419 \div 2 \times 3$] = 0.0789 moles of ${O}_{2}$.
Note: This answer is rounded to 3 significant figures.

Sorry if my explanation is too long or not concise, but I'm still trying my best so it makes sense to you.