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How many molecules of oxygen are produced when 6.54 g of potassium chlorite decompose?

User Jlbriggs
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Answer:

3.24× 10^22 molecules of oxygen gas.

Step-by-step explanation:

The decomposition of potassium chlorite produces oxygen gas according to the following equation:

2KClO2(s) → 2KCl(s) + 3O2(g)

From the equation, we can see that for every 2 moles of KClO2 that decompose, 3 moles of O2 are produced.

To determine the number of moles of KClO2 in 6.54 g, we need to divide the mass by the molar mass of KClO2. The molar mass of KClO2 is

121.56 g/mol (39.10 g/mol for potassium + 35.45 g/mol for chlorine + 47.01 g/mol for oxygen).

moles of KClO2 = 6.54 g / 121.56 g/mol = 0.0538 mol

Since 2 moles of KClO2 produce 3 moles of O2, we can use this ratio to find the number of moles of O2 produced:

moles of O2 = 0.0538 mol KClO2 × (3 mol O2 / 2 mol KClO2) = 0.08mol O2

Finally, we can use Avogadro's number, which is 6.022 × 10^23 molecules/mol, to convert the number of moles of O2 to the number of molecules of O2:

number of molecules of O2 = 0.0538mol O2 × 6.022 × 10^23 molecules/mol = 3.24× 10^22 molecules of O2

Therefore, 6.54 g of potassium chlorite produces approximately 3.24× 10^22 molecules of oxygen gas.

User Mahsa
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