Question

23 of 33 Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3(s).KClO3(s). The equation for the reaction is 2KClO3⟶2KCl+3O2 2KClO3⟶2KCl+3O2 Calculate how many grams of O2(g)O2(g) can be produced from heating 49.7 g KClO3(s).

Answer 1

19.488g

Step-by-step explanation:

Now, we first need to know the mole ratio of the molecules we are dealing with. From the equation of reaction, we can see that 2 moles of KClO3 yielded 3 moles of oxygen gas. Let's keep this as it would be useful later on.

The above paragraph illustrates what is expected and does not say what is actually obtained I.e the reacting number of moles.

The number of moles is related to the mass by dividing the mass by the relative molecular mass. Thus, we can calculate the actual number of moles of KClO3 reacted by dividing the mass by the relative molecular mass.

The relative molecular mass is calculated as 39 + 35.5 + 3(16) = 122.5g/mole.

Hence the number of moles of KClO3 reacted would be 49.7 ÷ 122.5 = 0.406 moles

Now, we go back to our first paragraph.

If 2 moles of KClO3 yielded 3 moles of O2, then 0.406 KClO3 will yield (0.406 × 3) ÷ 2 = 0.609 moles 02

We can now from here calculate the mass of oxygen produced.

The mass of oxygen produced = number of moles of oxygen yielded × relative molecular mass of oxygen.

The relative molecular mass of oxygen gas is 32g/mole.

Hence the mass would be = 32 × 0.609 = 19.488g

Answer 2

20 g O₂

Step-by-step explanation:

To solve this question we will be using the stoichiometry of the balanced equation . In order to do that, lets first convert the quantity of 49.7 g KClO3 in mol and then we will do the calculations to solve our question.

MW KClO3 = 122.55 g/mol

49.7 g x 1 mol/122.55 g / mol = 0.41 mol

The stoichiometry of the reaction tells us that 3 moles of O2 are produced from 2 molles of KClO3 , then we can calculate the moes from 0.41 of KClO3 and convert it to grams and answer our question.

0.41 mol KClO₃ x 3 mol O₂/ 2 mol KClO₃ = 0.61 mol O₂

= 0.61 mol O₂ x 32 g //mol O₂ = 20 g O₂