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

Answer:

$m_{O_2}=21.7gO_2$

Explanation:

Hello,

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

$m_{O_2}=55.3gKClO_3*\frac{1molKClO_3}{122.55 gKClO_3}*\frac{3molO_2}{2molKClO_3}*\frac{32gO_2}{1molO_2} \\m_{O_2}=21.7gO_2$

Best regards.

Answer 2

Answer:

21.6 grams of O2 can be produced

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