Question

The volume of oxygen obtained when 9.60g of ozone undergoes decomposition at STP is

Answer 1

Answer:

4.48L

Explanation:

given : 9.60g O3

moles of O3 9.60/48=0.2mol

1 mol O3 gives 22.4L O2

then

0.2 mol O3 gives 22.4*0.2=4.48L

Answer 2

The volume of oxygen gas obtained is 6.72 L

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of ozone = 9.60 g

Molar mass of ozone = 48 g/mol

Putting values in above equation, we get:

$\text{Moles of ozone}=\frac{9.60g}{48g/mol}=0.2mol$

The chemical equation for the decomposition of ozone follows:

$2O_3\rightarrow 3O_2$

By Stoichiometry of the reaction:

2 moles of ozone produces 3 moles of oxygen gas

So, 0.2 moles of ozone will produce = $\frac{3}{2}\times 0.2=0.3mol$ of oxygen gas

At STP:

1 mole of a gas occupies 22.4 L of volume

So, 0.3 moles of oxygen gas will occupy = $\frac{22.4}{1}\times 0.3=6.72L$ of volume

Learn more about number of moles and volume:

https://brainly.com/question/919137

https://brainly.com/question/13972001

#learnwithbrainly