Question

2) 2KClO3 --> 2KCl + 3O2

Answer 1

$2 \text{ KClO}_3 \to 2 \text{ KCl}+3\text{ O}_2$

a)

$2 \text{ mols of KClO}_3 \equiv 3 \text{ mols of O}_2$

$19 \text{ mols of KClO}_3 \equiv 3\cdot 9,5 \text{ mols of O}_2$

$\boxed{19 \text{ mols of KClO}_3 \equiv 28,5 \text{ mols of O}_2}$

b)

$2 \text{ mols of KClO}_3 \equiv 2 \text{ mols of KCl}$

$62 \text{ mol of KClO}_3 \equiv 62 \text{ mol of KCl}$

Using the atomic mass given in the periodic table:

$62\cdot(39+35,5+16\cdot3) \text{ g of KClO}_3 \equiv 62 \text{ mol of KCl}$

$62\cdot122,5 \text{ g of KClO}_3 \equiv 62 \text{ mol of KCl}$

$7595 \text{ g of KClO}_3 \equiv 62 \text{ mol of KCl}$

$\boxed{7,595 \text{ kg of KClO}_3 \equiv 62 \text{ mol of KCl}}$

c)

$2 \text{ KCl}+3\text{ O}_2\to 2 \text{ KClO}_3$

$3 \text{ mols of O}_2 \equiv 2 \text{ mols of KCl}$

Using the atomic mass given in the periodic table:

$3\cdot(2\cdot 16) \text{ g of O}_2 \equiv 2\cdot(39+35,5) \text{ g of KCl}$

$96\text{ g of O}_2 \equiv 149\text{ g of KCl}$

$\dfrac{39}{149}\cdot 96\text{ g of O}_2 \equiv \dfrac{39}{149}\cdot 149\text{ g of KCl}$

$\boxed{25,13\text{ g of O}_2 \equiv 39\text{ g of KCl}}$

This result is an aproximation.