# What volume of oxygen is produced at STP when 6.58 times 10^24 molecules of water is decomposed?

Feb 5, 2017

Every two molecules of water make for one molecule of oxygen.

#### Explanation:

$2 {H}_{2} O \to 2 {H}_{2} + {O}_{2}$

So there will be $0.5 \times 6.58 \times {10}^{24} = 3.29 \times {10}^{24}$ ${O}_{2}$ molecules.

$1 m o l = 6.02 \times {10}^{23}$ particles, so the amount produced is:

$\frac{3.29 \times {10}^{24}}{6.02 \times {10}^{23}} = 5.47 m o l$

At an STP of $273 K , 1 a t m$ the volume is $22.4 L / m o l$
(you can look up molar volumes for any other STP, there are lots of STP's)

So the volume of oxygen is $5.47 \times 22.4 = 122 L$