Question

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

a) How many moles of O2 are produced from 19 moles of KClO3?
b) How many kilograms of KClO3 would decompose to form 62 moles of KCl?
c) How many grams of O2 are required to react with 39 grams of KCl?

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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}`

`(39)/(149)\cdot 96\text{ g of O}_2 \equiv (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.