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.

it's filtration, but this is an assumption so my apologies if i got this

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