Question

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3(s). The equation for the reaction is 2KClO3⟶2KCl+3O2 Calculate how many grams of O2(g) can be produced from heating 55.3 g KClO3(s). mass: O 2.

Answer 1

21.6 grams of O2 can be produced

Step-by-step explanation:

Step 1: Data given

Mass of KClO3 = 55.3 grams

Molar mass KClO3 = 122.55 g/mol

Molar mass O2 = 32.0 g/mol

Step 2: The balanced equation

2KClO3⟶2KCl+3O2

Step 3: Calculate moles KClO3

Moles KClO3 = mass KClO3 / molar mass KClO3

Moles KClO3 = 55.3 grams / 122.55 g/mol

Moles KClO3 = 0.451 moles

Step 4: Calculate moles O2

For 2 moles KClO3 we'll have 2 moles KCl and 3 moles O2

For 0.451 moles KClO3 we'll have 3/2 * 0.451 moles = 0.6765 moles O2

Step 5: Calculate mass O2

Mass O2 = moles O2 * molar mass O2

Mass O2 = 0.6765 moles * 32.0 g/mol

MAss O2 = 21.6 grams

21.6 grams of O2 can be produced

Answer 2

`m_(O_2)=21.7gO_2`

Step-by-step explanation:

Hello,

In this case, with the given chemical reaction, the stoichiometry applies for the resulting mass of oxygen as shown below:

`m_(O_2)=55.3gKClO_3*(1molKClO_3)/(122.55 gKClO_3)*(3molO_2)/(2molKClO_3)*(32gO_2)/(1molO_2) \\m_(O_2)=21.7gO_2`

Best regards.