How many grams of ko2 are needed to form 8.0 g of o2?

23.43 grams of KO₂ is needed to form 8.0 g of O₂

Stochiometry in Chemistry learns about chemical reactions mainly emphasizing quantitative, such as calculations of volume, mass, number, which are related to the number of actions, molecules, elements, etc.

In chemical calculations, the reaction can be determined, the number of substances that can be expressed in units of mass, volume, mole, or determine a chemical formula, for example, the substance level or molecular formula of the hydrate.

The reaction equation is the chemical formula of reagents and product substances

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

In the reaction equation there are also manifestations of reagent substances namely gas (g), liquid (liquid / l), solid (solid / s) and solution (aqueous / aq).

The reaction between potassium superoxide, KO₂, and CO₂, to get O₂, is usually used by rescue teams in the form of breathing equipment

4KO₂ + 2CO₂ → 2K₂CO₃ + 3O₂

From the above equation the reaction shows that for every 4 moles of KO₂ you will get 3 moles of O₂, or a ratio of moles of KO₂: moles of O₂ = 4: 3

If we want to get 8 grams of O₂, then the number of moles of O₂ becomes:

mole O₂ = gram: molecular mass of O₂

mol O₂ = 8 gram: 32

mol O₂ = 0.25

so the number of KO₂ moles needed: (we use a ratio of KO₂ and O₂ moles)

mol KO₂ = 4/3 x mol O₂

KO KO2 = 4/3 x 0.25

mol KO2 = 0.33 mol

So the number of grams of KO₂ needed

gram KO₂ = mol x molecular mass KO₂

gram KO₂ = 0.33 x 71

gram KO₂ = 23.43

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Keywords: mole, KO₂, O₂, breathing equipment

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