What volume of o2 collected at 22.0 and 728 mmhg would be produce by the decomposition of 8.15 g kclo3?

The volume of O₂ collected from the reaction is 2.52 L

We'll begin by calculating the number of mole in 8.15 g of KClO₃

Mole = mass / molar mass

Mole of KClO₃ = 8.15 / 122.5

Mole of KClO₃ = 0.0665 mole

Next, we shall determine the number of mole of O₂ produced from the reaction. This can be obtained as follow:

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Therefore,

0.0665 mole of KClO₃ will decompose to produce = (0.0665 × 3) / 2 = 0.09975 mole of O₂.

Finally, we shall determine the volume of O₂.

Using the ideal gas equation, we can obtain the volume of the O₂ as follow:

PV = nRT

0.958 × V = 0.09975 × 0.0821 × 295

Divide both side by 0.958

V = (0.09975 × 0.0821 × 295) / 0.958

V = 2.52 L

Thus, the volume of O₂ obtained from the reaction is 2.52 L

Learn more about ideal gas equation:

link

There is 2.52 L of O2 collected

Explanation:

Step 1: Data given:

Temperature = 22.0 °C

Pressure = 728 mmHg = 728 /760 = 0.958 atm

Mass of KClO3 = 8.15 grams

Molar mass of KClO3 = 122.55 g/mol

Step 2: The balanced equation

2KClO3(s) → 2KCl(s) + 3O2(g)

Step 3: Calculate moles of KClO3

Moles KClO3 = mass KClO3 / molar mass KClO3

Moles KClO3= 8.15 grams / 122.55 g/mol

Moles KClO3 = 0.0665 moles

Step 4: Calculate moles of O2

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

For 0.0665 moles of KClO3 we have 3/2 * 0.0665 = 0.09975 moles

Step 5: Calculate vlume of O2

p*V = n*R*T

V = (n*R*T)/p

⇒ with n = the number of moles O2 = 0.09975 moles

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = 22.0 °C = 273 +22 = 295 Kelvin

⇒ with p = 0.958 atm

V = (0.09975 * 0.08206 * 295) / 0.958

V = 2.52 L

There is 2.52 L of O2 collected

the volume increases