[tex]2 \text{ KClO}_3 \to 2 \text{ KCl}+3\text{ O}_2[/tex]
a)
[tex]2 \text{ mols of KClO}_3 \equiv 3 \text{ mols of O}_2[/tex]
[tex]19 \text{ mols of KClO}_3 \equiv 3\cdot 9,5 \text{ mols of O}_2[/tex]
[tex]\boxed{19 \text{ mols of KClO}_3 \equiv 28,5 \text{ mols of O}_2}[/tex]
b)
[tex]2 \text{ mols of KClO}_3 \equiv 2 \text{ mols of KCl}[/tex]
[tex]62 \text{ mol of KClO}_3 \equiv 62 \text{ mol of KCl}[/tex]
Using the atomic mass given in the periodic table:
[tex]62\cdot(39+35,5+16\cdot3) \text{ g of KClO}_3 \equiv 62 \text{ mol of KCl}[/tex]
[tex]62\cdot122,5 \text{ g of KClO}_3 \equiv 62 \text{ mol of KCl}[/tex]
[tex]7595 \text{ g of KClO}_3 \equiv 62 \text{ mol of KCl}[/tex]
[tex]\boxed{7,595 \text{ kg of KClO}_3 \equiv 62 \text{ mol of KCl}}[/tex]
c)
[tex]2 \text{ KCl}+3\text{ O}_2\to 2 \text{ KClO}_3[/tex]
[tex]3 \text{ mols of O}_2 \equiv 2 \text{ mols of KCl}[/tex]
Using the atomic mass given in the periodic table:
[tex]3\cdot(2\cdot 16) \text{ g of O}_2 \equiv 2\cdot(39+35,5) \text{ g of KCl}[/tex]
[tex]96\text{ g of O}_2 \equiv 149\text{ g of KCl}[/tex]
[tex]\dfrac{39}{149}\cdot 96\text{ g of O}_2 \equiv \dfrac{39}{149}\cdot 149\text{ g of KCl}[/tex]
[tex]\boxed{25,13\text{ g of O}_2 \equiv 39\text{ g of KCl}}[/tex]
This result is an aproximation.
