Answer:
6.84L of O2 can be produced.
Explanation:
1st) It is necessary to write and balance the chemical reaction:
[tex]2KClO_3\rightarrow2KCl+3O_2[/tex]
From the balanced reaction we know that 2 moles of KClO3 decompose into 2 moles of KCl and 3 moles of O2.
2nd) It is necessary to calculate the molar mass of KClO3 and O2:
- KClO3 molar mass: 122.6g/mol
- O2 molar mass: 32g/mol
3rd) In the conversion we have to consider:
- KClO3 molar mass.
- The stoichiometry of the reaction between KClO3 and O2: from 2 moles of KClO3, 3 moles of O2 are produced.
- The relation of moles and volume: 1 mole occupies 1 L.
[tex]24.95gKClO_3*\frac{1mol}{122.6g}*\frac{3molesO_2}{2molesKClO_3}*\frac{22.4L}{1moleO_2}=6.84L\text{ }O_2[/tex]
So, 6.84L of O2 can be produced.
