Question

in the following equation 2KCIO3 to 2KCI + 3O2 HOW MAY MOLES OF OXEGEN, O2 are produced by thee decomposition to 1.05 grams of potassium cholorate KCIO3

Answer 1

Explanation:

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Answer 2

Answer:

0.013 moles of O₂ are produced by the decomposition to 1.05 grams of potassium cholorate KCIO₃

Explanation:

You know the reaction:

2 KCIO₃ → 2 KCI + 3 O₂

First of all you need to know the molar mass of each compound that participates in the reaction. For that you get the atomic mass of the elements that make up each compound in the Periodic Table of the elements:

• K: 39 g/mole
• Cl: 35.45 g/mole
• O: 16 g/mole

Then, the molar mass of the compounds involved in the reaction, taking into account the abundance of the elements that form the compounds and their atomic mass, is:

• KClO₃= 39 g/mole + 35.45 g/mole + 3*16 g/mole= 122.45 g/mole
• KCl= 39 g/mole + 35.45 g/mole= 74.45 g/mole
• O₂= 2*16 g/mole= 32 g/mole

In moles, by reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction) participate:

• KClO₃= 2 moles
• KCl= 2 moles
• O₂= 3 moles

Then, in mass, by stoichiometry of the reaction participate:

• KClO₃= 2 moles*122.45 g/mole= 244.9 g
• KCl= 2 moles*74.45 g/mole= 148.9 g
• O₂= 3 moles*32 g/mole= 96 g

Now you apply a rule of three as follows: if by stoichiometry 244.9 g of KClO₃ form 3 moles of O₂, 1.05 g of KClO₃ how many moles of O₂ will they form?

$moles of O_{2} =\frac{1.05 g of KClO_{3}*3 moles of O_{2} }{244.9 g of KClO_{3} }$

moles of O₂≅ 0.013

0.013 moles of O₂ are produced by the decomposition to 1.05 grams of potassium cholorate KCIO₃