How many moles of O2, form as 80 g of KClO3 are totally consumed?
M(KCIO3)= 122.45 g/mol.

0.975 mole of O₂

Explanation:

We'll begin by calculating the number of mole in 80 g of KClO₃. This can be obtained as follow:

Molar mass of KClO₃ = 122.45 g/mol.

Mass of KClO₃ = 80 g

Mole of KClO₃ =?

Mole = mass /Molar mass

Mole of KClO₃ = 80 / 122.45

Mole of KClO₃ = 0.65 mole

Next, we shall write the balanced equation for the reaction. This is illustrated below:

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Finally, we shall determine the number of mole O₂ produced by the decomposition of 80 g (i.e 0.65 mole) of KClO₃. This can be obtained as follow:

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Therefore, 0.65 mole of KClO₃ will decompose to produce = (0.65 × 3)/2 = 0.975 mole of O₂.

Thus, 0.975 mole of O₂ was obtained from the reaction.

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.