# In the equation 2KClO_3 -> 2KCl + 3O_2, if 5.0 g of KClO_3 is decomposed, what volume of O_2 is produced at STP?

Apr 3, 2016

#### Answer:

$1.371 L$

#### Explanation:

Find the relative mass of the molecule $K C l {O}_{3}$ by adding together the masses of each atom in it. Find the masses from the periodic table.

$39.01 g m o {l}^{-} 1 + 35.45 g m o {l}^{-} 1 + 3 \cdot 16.00 g m o {l}^{-} 1$
$= 122.46 g m o {l}^{-} 1$

Divide the total mass of the substance by its relative mass to find the number of moles of the substance.

$\frac{5.00 g}{122.46 g m o {l}^{-} 1} = 0.041 m o l$

There are two moles of $K C l {O}_{3}$ for every 3 moles of ${O}_{2}$, which we can see from the equation. Using the ratio $2 : 3$ and inserting $0.0408 m o l$, we find that there must be $0.0612 m o l$ of ${O}_{2}$.

One mole of an ideal gas at STP (standard temperature and pressure) occupies 22.4L.

Multiplying

$22.4 L m o {l}^{-} 1 \cdot 0.0612 m o l = 1.371 L$

which should be your final answer, depending on how you've rounded.