Solid potassium chlorate (KClO3)(KClO3) decomposes into potassium chloride and oxygen gas when heated. How many moles of oxygen form when 48.1 gg completely decomposes

0.59 mol O₂

Explanation:

The balanced chemical equation for the decomposition of potassium chlorate (KClO₃) to produce potassium chloride (KCl) and oxygen gas (O₂) is the following:

2 KClO₃ → 2 KCl + 3 O₂

According to the equation, 3 moles of O₂ are produced from 2 moles of KClO ⇒ conversion factor: 3 mol O₂/2 mol KClO₃

Now, we calculate the number of moles of KClO₃ there is in 48.1 g, by dividing the mass into the molecular weight (Mw) of O₂:

Mw(KClO₃) = 39.1 g/mol + 35.4 g/mol + (16 g/mol x 3) = 122.5 g/mol

moles KClO₃ = mass KClO₃/Mw(KClO₃) = 48.1 g/(122.5 g/mol) = 0.3926 mol KClO₃

Finally, we multiply the moles of KClO₃ by the conversion factor to calculate the moles of O₂ produced:

0.3926 mol KClO₃ x 3 mol O₂/2 mol KClO₃ = 0.59 mol O₂

the correct answer is:

a. formation of precipitation

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.