If an electric discharge produces 800 cm3 of ozone (O3), how many cm3 of oxygen (O2) are required?
3O2(g) ---> 2O3(g)
When gases are involved one may use stoichiometry as if volume = mols.
800 cc O3 x (3 mols O2/2 mols O3) = 800 x 3/2 = ? cc O2
To determine the amount of oxygen required, we can use the stoichiometric ratio expressed in the equation:
3O2(g) ---> 2O3(g)
From the equation, we can see that 3 moles of oxygen (O2) are required to produce 2 moles of ozone (O3).
1 mole of gas at standard temperature and pressure (STP) occupies 22.4 liters or 22,400 cm3. Therefore, 1 mole of any gas will occupy 22,400 cm3.
To solve the problem, we need to convert the given volume of ozone (800 cm3) to moles using the molar volume at STP:
800 cm3 * (1 mole / 22,400 cm3) = 0.0357 moles of ozone (O3)
Now, we can use the stoichiometric ratio to convert moles of ozone to moles of oxygen:
(0.0357 moles O3) * (3 moles O2 / 2 moles O3) = 0.0535 moles of oxygen (O2)
Finally, we can convert the moles of oxygen back to volume using the molar volume at STP:
0.0535 moles O2 * (22,400 cm3 / 1 mole) = 1,196 cm3 of oxygen (O2)
Therefore, approximately 1,196 cm3 of oxygen (O2) are required to produce 800 cm3 of ozone (O3) through electric discharge.