Question

How many grams of potassium chlorate must be heated to produce 30.0g of oxygen? socratic.org

Answer 1

g KCLO3 = 76.594 g

Step-by-step explanation:

• KClO3 → KCl + 3/2O2

∴ Mw KClO3 = 122.55 g/mol

∴ Mw O2 = 32 g/mol

⇒ g KClO3 = ( 30 g O2)×(mol O2/32 g)×(mol KClO3 / 3/2 mol O2)×(122.55 g KClO3/ mol)

⇒ g KCLO3 = 76.594 g

Answer 2

76.56g

Step-by-step explanation:

Firstly, to do this we need a correct and balanced equation for the decomposition of potassium chlorate.

2KClO3 —-> 2KCl + 3O2

From the balanced equation, we can see that 2 moles of potassium chlorate yielded 3 moles of oxygen gas

We need to know the actual number of moles of oxygen gas produced. To do this, we divide the mass of the oxygen gas by its molar mass. Its molar mass is 32g/mol

The number of moles is thus 30/32 = 0.9375 moles

Now we can calculate the number of moles of potassium chlorate decomposed.

We simply do this by (0.9375 * 2)/3 = 0.625 moles

Now to get the number of grammes of potassium chlorate decomposed, we simply multiply this number of moles by the molecular mass. The molecular mass of KClO3 is 39 + 35.5 + 3(16) = 122.5g/mol

The amount in grammes is thus 122.5 * 0.625 = 76.56g